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Full text of "Qualitative Testing And Inorganic Chemistry"


158107 >m 


Call No - Acc " No ' 

Osmania University Library 

Call No. S4-A- Accession No. 



This book^should be returned on or befoib the ds^te last 
marked below. 




There is a trend today to abandon the second-semester freshman 
course in qualitative analysis in favor of a year's general chemistry 
which includes a few weeks of analysis and chemical equilibrium. I 
think that some excellent chemistry, particularly in the laboratory, is 
lost that way, and in this book I use the opposite approach to cur- 
riculum revision by emphasizing the best of the "qual" and adding 
related general chemistry to that. Modernization of the stereotyped 
"qual" program is desirable, because it has always seemed to admit 
its laboratory procedures have little value outside the classroom and 
palmed off the theoretical sections as chemical major preparation for 
advanced study. I never became enthusiastic over that, neither did 
the nonmajors who constitute the large majority of most classes, since 
they were not getting anything they could use. As this is the last 
chemistry required of most people, one should not feel obligated to 
fit time-honored material into a narrow, half-century old outline. I 
therefore try to develop the subject in both lecture and laboratory as 
something alive with applications for any physical science student by 
presenting it not as an isolated factual body but as a relative of many 
scientific areas. This makes for a rather large volume although not 
necessarily one written down to a descriptive level. The text probably 
cannot be completely covered in a semester, but the instructor can 
vary the course within it, and extra reading and laboratory work are 
available as needed. 

This book assumes a reader's sound background of at least one 
semester or two quarters of general college chemistry which considered 
such topics as properties of matter, chemical symbolism, nomen- 



clature and equations, structure of atoms and molecules and their 
weight relationships, the periodic system, the gas laws, types of solu- 
tions, introductory redox and equilibrium, and descriptive chemistry 
of important nonmetals. The second course may then be free to use 
its own text, such as this one, for the chemistry of the metals, some 
topics in "modern" inorganic chemistry, the elaboration of equilibrium 
systems, and the qualitative testing of selected cations and anions. To 
tie these topics together I use examples and problems from the labora- 
tory work and industry and explain how our knowledge developed 
historically through experimentation. The emphasis is on the broad 
view of inorganic chemistry. 

After a review of equation writing and calculations on solution con- 
centration (Chapters 1 and 2), atomic structure and bonding are re- 
viewed and extended for use in explaining the constitution of Werner 
ions and, together with conductance studies, in formulating the acid- 
base theories (Chapters 3, 4, and 5). The next six chapters develop 
chemical equilibrium with reference to substances handled in the labora- 
tory, and many experimental methods for determining equilibrium 
constants are explained to give the reader more feeling for the ab- 
stractions. Chapter 12 classifies some organic qualitative reagents. 
The next chapter briefly treats process metallurgy and the crystalline 
structure of metals. Chapter 14 outlines general laboratory procedures, 
and the following five chapters cover the usual five cation groups. 
These include a factual description of the chemistry of each metal, an 
explanation and flowsheet of group analysis, preliminary laboratory 
study of the ions, detailed procedure for sample analysis, and all re- 
action equations. The preliminaries are almost entirely alternate tests 
for ions rather than a duplication of descriptive material. It is be- 
lieved this will allow more analytical time on samples, yet many of the 
preliminaries will be done conscientiously in the course of "unknown" in- 
vestigation, since the checking of the sample using blanks, "knowns," 
and confirmatory examinations has added meaning and incentive. 

Chapter 20 describes eight "new" metals, Li, Be, Ti, Zr, V, Mo, W, 
and U, and contains enough laboratory direction for their identification 
in the usual alloys and matrixes. They are not included in the next 
chapter on the general cation unknown, however, although two of in- 
dustry's procedures are reprinted for identifying some commercial al- 
loys. Chapter 22 concerns the chemistry and analysis of 18 anions, 
mainly by spot testing so any or all can be included without changing 
laboratory directions. The last chapter is a collection of 13 so-called 
special experiments which can be employed as class demonstrations or 


assigned to individuals in place of more complex samples. A few 
of them illustrate the basis for theoretical principles, and the rest hold 
human and applied interest and demonstrate such qualitative testing 
techniques as microscopy, chromatography, fluorescence, electrography, 

Most chapters end with a set of problems, some with answers, and 
recent references to /. Chem. Educ. and Anal. Chem. The latter are 
only meant to be a literature introduction which interested students 
can read with benefit, and they are not meant to be research-type source 

The class-room instruction of qualitative analysis and inorganic chem- 
istry inherently includes the well-known problem of laboratory and 
lecture coordination. In the past, I used the lecture hours at the 
semester's beginning to rapidly review solution concentrations, equation 
writing, and chemical bonding (Chapters 1, 2, and 3) and used the 
quiz hours to give laboratory orientation and directions for group 1 and 
2 cations (Chapters 14, 15, and 16). After that I took up Werner 
compounds, acid-base theories and reaction rates (Chapters 4, 5, and 
6) and then started equilibrium (Chapters 7, 8, 9, 10, and 11). Other 
years I have varied this by beginning the equilibrium sections right after 
a review of the first two chapters so they coincide with the current lab- 
oratory work: sulfide solubilities, pH control, etc. I then alternated 
descriptive chapters with those on calculations, for the sake of variety. 
The chapter on metallurgy is appropriate when a metal or mineral 
sample is analyzed, and the section on organic reagents can follow 
anion analysis or precede a general unknown. The special experiments 
can be used individually or as class demonstrations and interwoven 
wherever they seem indicated, for instance, flame and bead tests with 
cation groups 4 and 5. I usually finished the semester with several lec- 
tures on a topic of special interest, such as electrochemical corrosion, 
physical metallurgy, crystallography, or refractory metals and had fun 
with related industrial samples possessing analytical aspects, or on field 
trips to plants using metals and quality control. 

Whatever the instructor's preference for topics, presentation order, or 
his group's training and ability, I hope the material is here for a bene- 
ficial and interesting time. 

This book could not have been published without the splendid coop- 
eration and suggestions by the following people who gave generously 
of their valuable time and experience: Professors E. L. Haenisch, Wa- 
bash College, W. K. Wilmarth, University of Southern California, 


R. Nelson Smith, Pomona College, E. S. Kuljian, Pierce College, A. E. 
Flanigan, University of California at Los Angeles. 

Three former students of mine who helped check some of the labora- 
tory procedures and printer's proof, also deserve special mention : Linda 
Day, Richard Geer, and Leon Simon. 

Mrs. Eleanor Lovelee typed every word between these covers at least 


Van Nuys, California 
September, 1957 



1. Chemical Equations 1 

2. Expressing the Concentration of Solutions 18 

3. Electrons, Atoms, Bonds, and Molecules 28 

4. Werner Ions and Complex Compounds 67 

5. Acid-Base Theories and the Electrical Conductance of 
Electrolytes 82 

6. Reaction Rates and Chemical Equilibrium 96 

7. Equilibrium and Weak Acids and Bases 108 

8. Equilibrium and Slightly Soluble Electrolytes 127 

9. Equilibrium and Werner Tons 144 

10. Equilibrium and Hydrolysis 154 

11. Equilibrium and Redox: Electrochemical Cells 169 

12. Organic Reagents for Qualitative Analysis 190 

13. Minerals, Metals, and Crystals 199 

14. Orientation for the Laboratory Work 216 

15. Cation Group 1 231 

16. Cation Group 2 247 

17. Cation Group 3 



Chapter Page 

18. Cation Group 4 306 

19. Cation Group 5 319 

20. Some Less Familiar Engineering Metals 334 

21. Analysis of a General Cation Unknown 358 

22. The Anion Groups 374 

23. Special Experiments 417 
Appendix 449 
Index 477 



Few chemistry courses contain so many reactions of compounds of 
different elements as does qualitative analysis, and indeed one purpose 
of teaching this subject is to acquaint students with inorganic reactions. 
It becomes necessary, therefore, that one be able to write and balance 
chemical equations, which are the shorthand statements of reactions. 

It is assumed the student can write formulas and give names for simple 
compounds and that some of this chapter will be review material. Tables 
A24 and A25, giving valence numbers, symbols for elements, and formulas 
for radicals, will be helpful. 

Information That Equations Can impart 

1. Can give a simple description of reaction and the ratio of substances 

4Fe + 3O 2 = 2Fe 2 O 3 

This means that reaction between iron and oxygen produces ferric oxide. 
Each symbol like Fe and O represents one atom of a particular element, 
or Avogadro^s number (N) of atoms, 6.02 x 10 23 , which is also called a 
mole of particles. Each formula like Fe 2 O 3 stands for a molecule of a 
compound (whose composition is known by chemical analysis) or Avo- 
gadro's number of molecules, 6.02 X 10 23 . The coefficients 4, 3, and 2 
indicate that 4 atoms of iron and 3 molecules of oxygen combine to 
form 2 molecules of ferric oxide. The subscripts 2, 2, and 3 indicate, 



respectively, that the oxygen molecule is composed of 2 atoms of oxygen 
and that the Fe 2 O 3 molecule is made of 2 atoms of iron and 3 of oxygen. 
A coefficient multiplies each atom and subscript in the entire term it 
precedes; a subscript multiplies only the atom or radical that precedes it. 
The principal reaction is conventionally the one pictured going from 
left to right. The left-side terms are called reactants or starting materials, 
and the terms on the right side are products or resultants. 

2. Can give the weight ratios (and volume ratios for gases) of substances 
present. Since each term may stand for N particles, and since TV atoms com- 
prise digram atomic weight and Wmolecules a gram molecular weight, one may 
use the properly written equation to make calculations in what are known 
as weight-weight, weight-volume, volume-volume, and material-balance 
problems. A mole of oxygen molecules occupies 22.4 liters, a volume 
known as a gram molecular or molar volume, at standard conditions (s.c.) 
of temperature (0 C = 273 A) and pressure (760 mm Hg = 13.7 Ib/in. 2 
= 1.00fl/w). The equation therefore gives the information that 
(4)(55.85) g Fe + (3)(32.00) g O 2 or (3)(22.4) liters O 2 react to produce 
(2)[(2X55.85) + (3)(32.00)J g of Fe 2 O 3 if the process is 100% complete. 
This maximum quantity is called the theoretical yield of the chemical 
process. Grams and liters are the laboratory units most used. Engi- 
neering units are usually pounds and cubic feet. A pound molecular 
gas volume at standard conditions is 359 cubic feet. The total weights 
of reactants must equal the total weights of products in ordinary (non- 
nuclear energy) reactions. 

3. Can show the physical states of the substances. It will be the con- 
vention for equations in this book to represent solids in bold face, gases 
in italic, and dissolved substances and liquids in ordinary type. For the 
student writing equations, solids may be underlined and precipitates 
(Solubility Table, back cover) may be accompanied by a downward 
pointing arrow, as AgCl | ; gases may have a line over them and/or an 
upward arrow, as NO 2 1 . Alternately, the symbols (g), (/), (s) indicating 
gas, liquid, solid are also used. For example H 2 O(^) means steam or 
water vapor. Strong acids, strong bases, and most salts dissociate to 
yield ions in aqueous solution, whereas weak acids (Table A20), weak 
bases (Table A21), and a few heavy metal salts are weakly ionized and 
accordingly represented as molecules. Oxidation states of elements and 
valences of radicals will be shown in equations as needed (see the sections 
on balancing redox reactions). Water will be shown if it is a reactant or 
is necessary to clarify the mechanism, but it is otherwise omitted as most 
reactions in this course are understood to be in aqueous solution. The 

H+ + HSO^ + Pb = 7/ 2 + PbSO 4 


accordingly means that sulfuric acid dissociates in steps (the first of 
which is complete) and reacts with metallic lead to produce hydrogen 
gas and a precipitate of lead sulfate. 

4. Can indicate reversibility. For many purposes one may say that a 
reaction is essentially complete in the direction in which a gas is formed 
(and escapes), a precipitate comes down, or a slightly ionized compound 
is produced (H 2 O, weak acids, weak bases, some heavy metal salts). Since 
even these processes are not 100% conversions, an equals sign or double 
arrows is used to show a state of reversibility or equilibrium in which at 
least minute amounts of each substance are present, and a heavier or 
longer arrow may be used to show the direction in which most of the 
material is displaced. 

Ag+ + Cl- ^ AgCl 

This is interpreted to mean that the silver ion and chloride ion react in 
water to precipitate silver chloride, but AgCl, not being completely 
insoluble, is in equilibrium with a low concentration of its ions. The 
compounds listed in tables in Chapters 15-20 are some common ones 
with low aqueous solubility. If the condition of reversibility of a reaction 
is not known, the equals sign may be used. A single arrow will mean 
the reaction is for all purposes complete as 

5. Can describe special features of the reaction. An exothermic reaction 
evolves heat and an endothermic reaction absorbs heat. Heat of reaction 
will be designated accordingly H r or - H r . If the actual value of H r is 
known or is of importance it is given in kilocalories per mole of product. 
If heat is externally applied to a system to cause it to proceed, a triangle, 
or actual temperature, will be put with the arrow or equals sign. If one 
is discussing the rates or speeds of the reaction to the right or left, the 
symbols S T and s l or s l and s 2 may be used with their respective arrows, 
and rate constants ^ and k 2 will be handled the same way (see Chapter 
6). If no reaction takes place, the letters N.R. or the words "no reaction," 
will be used where products would appear. Catalysts will also be indicated 
with the equals sign. 

Some reactions are photochemical processes, that is they are dependent 
upon energy from light for initiation, and the symbol used for a quantum 
of light will be hv (Chapter 3). Mathematical symbols referred to in 
problem solving will often be written under the corresponding terms in 
the chemical equation to insure clear understanding of the substitutions. 


This symbolism is interpreted to mean that liquid hydrogen peroxide is 
decomposed by heating with a solid ferric oxide catalyst, yielding steam 
and oxygen gas. 

H 2 O + CO i H 2 + CO 2 - H r 


This means that steam and carbon monoxide react at rate s r to form 
hydrogen and carbon dioxide which themselves react at rate s l to return 
the original reactants. In going to the right, heat is absorbed. H r is 
the heat of reaction. 

6. Can give colors of substances. The system of abbreviations of colors 
to be used throughout the book is given in Table A27. Colors of precipi- 
tates and of ions in solution are helpful in identifying samples. Combina- 
tions of these abbreviations may be made as 

4Fe+i + 3Fe(CN)^ 4 = FeJFe(CN) 6 ] 3 ,, flM 

meaning 4 moles of yellow-orange ferric ions react with 3 moles of 
yellow ferrocyanide ions to precipitate a mole of dark-blue ferric ferro- 
cyanide molecules. 

7. Can show structures and bonding. Sometimes it is important to 
know how many electrons are available for reaction and how radicals 
and molecules are bonded together. Accurate representation of structural 
features often enables chemists to predict the course of reactions and the 
constitution of the products. Quite a number of examples of this are 
given in later chapters, particularly those on bond types, complex ions, 
and organic chemistry. 

Balancing Simpler Equations 

The first step in equation writing is to give correct symbols and/or 
formulas for all reactants and products. The next step is to establish by 
means of coefficients the same number of like atoms on both sides, and, if 
charged particles are shown, equivalent total charge on both sides. For 
example, aluminum metal will react with silver ions to give aluminum 
ions plus silver metal. If one ignores a balance with respect to electrical 
neutrality, then the equation 

Al + Ag+ = A1+ 3 + Ag 

is incorrect, although correct symbols for all substances are shown and 
the numbers of atoms balance. Noting that the left side must have 3 
positive charges to balance the 3 on the right side, one puts the coefficient 


3 in front of Ag+; this means that 3 will be used as a coefficient for Ag 
on the right side for material balance: 

AI + 3Ag+ = A1+ 3 + 3Ag 

Predicting the Course of Simpler Reactions 

In order to write equations one must have firsthand knowledge of the 
chemical characteristics of the reactants or be able to predict them by the 
aid of some generalizations. Equations should not only be balanced but 
should also have the possibility of actually proceeding as written. This 
guarantee of reaction cannot always be given, of course, so research work 
often calls for analysis of reaction mixtures to see what happened. 

In general, reactions go essentially "to completion" if (a) a gas is 
formed (b) an insoluble compound precipitates (see Table A22 and the 
table inside the back cover) (c) a weakly ionized substance forms (Tables 
A20, A21, A22, and a few heavy metal salts mentioned in the descriptive 
sections on the ions), and (d) the emfofa redox reaction is positive (Tables 
A18, A19, and discussions in this chapter and Chapter 11). These four 
principles will always be useful to remember, though they do not always 
aid in predicting what the reaction products will be. For this more 
specific information is needed. 

Chemically similar substances are expected to react alike, and the 
following summary of type reactions is derived from such comparisons. 
With this aid, plus the periodic table (Chapter 3), the chemical activity 
series (this chapter), and the tables mentioned above, one can correctly 
predict the outcome of many of the simpler reactions. 

Example 1-1. In the group 1 cation reactions (Chapter 15) one mixes 
solns.* of potassium chromate and lead nitrate and a yellow ppt. is observed 
to form. What is the equation ? 

Following the type reactions summary, one recognizes the reactants as two 
salts and expects, according to reaction (3) under IV, (the numbering is purely 
arbitrary), two other salts to form. Each can only form from combination of 
a + to a ion, so there is only one other combination of the four ions possible. 
The correct formulas for reactants and products are 

K 2 Cr0 4 + Pb(N0 3 ) 2 = KNO 3 + PbCrO 4 

The left side has 2 K's and 2 NO 3 's which must be accounted for by putting 
the coefficient 2 in front of KNO 3 . This balances the molecular equation: 

K 2 Cr0 4 + Pb(N0 3 ) 2 = 2KNO 3 + PbCrO 4 

Inspection of the solubility table shows that of these, only lead chromate is 
insoluble, so the other ions are merely present as spectators in soln. and do not 

* Abbreviations are given in the Appendix, Table A26. 


participate in the essential reaction. Stripped of them, the ionic equation is 

Pb+ 2 + CrOj 2 = PbCrO 4F 

Example 1-2. What happens when a strip of silver and a strip of zinc are 
put in a cupric sulfate soln. ? 

Under (1) of III in the type reactions, one expects a single replacement reaction 
because the reactants are a metal and a salt. Checking the positions of Ag, 
Cu, and Zn in Table 1-1 shows Ag to be less easily oxidized than Cu, so Ag 
will not react with Cu +2 , but Zn which is above Cu and more readily oxidized 
will react. The salt formed will be zinc sulfate, but since it is ionic and sol., 
the essentials of the reaction mixt. are, 

Zn + Cu+ 2 = Zn+ 2 + Cu 

Example 1-3. A sodium thiosulfate soln. is acidified with hydrochloric 
acid. The mixt. bubbles, a sharp smelling gas is evolved, and the soln. turns 
milky. Explain. 

This is an example of type (4) under IV. It is noted there that as thiosulfuric 
acid is displaced by a strong acid, it decomp. to give sulfur dioxide (the gas) 
and sulfur (the milky precipitate). If the soln. is evaporated, NaCl would 
be found, but in soln., Na+ and Cl~ do not affect the reaction course so the 
process is 

2W + S 2 0i a = (H 2 S 2 3 ) = H 2 + S0 2 + S 

The parenthesis is used to denote an unstable intermediate. 

The following reaction types are a useful starting point for predicting 
reaction products, but the student must realize that there are some exceptions 
to these simple rules. 


I. Simple Syntheses 
A. Two Elements = Compound 

*(1) Metal + O 2 = Metal oxide 

Exceptions: Au and Pt give no reaction; Na, K, Ba give peroxides. 
Higher oxidation states of metal are usually formed. 

*(2) Nonmetal -f O 2 = Nonmetal oxide 

Exceptions: Periodic table group O (rare gases) give no reaction. 
*(3) Metal + S = Metal sulfide 

Lower reaction temperatures favor lower oxidation states of metal. 
* More important types among qualitative analysis reactions. 


*(4) Metal + X 2 = Metal halide (X 2 = halogen, periodic group VI I A) 

Higher oxidation states of metal expected with light halogens. I 2 often 
gives low yields. 

(5) Very active metals + H 2 = Metal hydride (H is the anion) 
Very active metals are the first 8 in Table 1-1. 





Ease of 

Not re- 

Not re- 




ing re- 



duced by 

duced by 

sis of 









occur in 




















by car- 

sis of 









by hydro- 










React directly 


and com- 

with oxygen to 



form oxides 







3y heat 

sis or 



Oxides are 






Related tables are A18 and A19. 


(6) Active nonmetals + H 2 = Nonmetal hydride 

A catalyst may be needed, as in NH 3 synthesis; others, like HF or HC1 
formation, may be explosive at low temperature. 

B. Element + Compound = More Complex Compound 

(1) Sulfite + 2 = Sulfate 

(2) Nitrite + O 2 = Nitrate 

C. Compound A + Compound B = More Complex Compound 
*(1) Acidic oxide + H 2 O = Acid 

(CO 2 -> H 2 CO 3 , SO 2 -> H 2 SO 3 , SO 3 -> H 2 SO 4 , N 2 O 3 -> HNO 2 , 
N 2 6 -*HN0 3 , P 4 6 ->H 3 P0 3 , P 4 10 ->H 3 P0 4 , C1 2 O 5 ^HC1O 3 , etc.) 
*(2) Basic oxide -f H 2 O = Base (some are hydrated oxides) 
*(3) Salt + NH 3 = Ammoniate (ammine) 

(See Chapter 4. Generally the coordination number is twice the metal 
oxidation number. Transition metal salts are common examples.) 

(4) Salt + H 2 O = Hydrate (One needs to learn specific examples.) 

(5) Basic oxide + Acidic oxide = Salt 

Metal oxides, except those of some high oxidation states like Mn 2 O 7 , 
are basic; nonmetal oxides including SiO 2 are acidic. 

II. Simple Decompositions 

These generally infer the reverse of the preceding syntheses. Decom- 
position may be effected by supplying heat, light, and/or electrical energy. 

Exceptions : Carbonates of periodic group IA (alkali metals) and of 
barium are not appreciably decomposed by heat. Chlorates give 
chlorides, not chlorites, upon heating. Only oxides and sulfides of 
metals quite low in the Electrochemical Series for Some Metallic Elements, 
Table 1-1, are easily decomposed by heat. The sulfur is oxidized in 
hot air to SO 2 gas. 

HI. Single Replacements 

A. Element A -f Compound A == Element B + Compound B 

*(1) Metal + salt = Metal' + salt' 

Reacting metal is above displaced metal (ion) in electrochemical series. 
Note which can be chemically displaced from solution (without electrolysis). 


*(2) Metal + acid = H 2 + salt 

See electrochemical series. Metals below H may react with acids giving 
H 2 O, a salt, and acid reduction products, but not H 2 gas. 

*(3) Halogen + halide = Halogen' + halide' 

Reacting elemental halogen has a smaller atomic number than the dis- 
placed halogen or reaction doesn't occur. 

(4) Metal + H 2 O = H 2 + oxide or hydroxide 

See electrochemical series; from Mg up, hydroxides are the products. 
Note the temperatures needed. 

(5) Metal oxides + C or 7/ 2 = Metal + CO, C0 2 , or H 2 O 

See electrochemical series. Metals easiest to reduce are at the bottom. 

(6) Active metals + base = H 2 + salt 

Metal is part of complex anion in products like Zn(OH)j 2 and Al(OH)^. 

IV. Double Replacements 

A. Compound A + Compound B = Two or More Other Com- 

*(1) Acid + base = Salt + H 2 O 

Acid of type HX, base of type MOH = Salt MX + H 2 O. 

*(2) Salt + H 2 = (Weak) acid + (weak) base 

Either or both products must be weakly dissociated for this hydrolysis to 
take place; see Chapter 10. 

*(3) Salt A + salt B = Salt C + salt D 

This proceeds if one or both products precipitates or is weakly ionized. 
See solubility data, back cover. 

*(4) Salt of weak acid + strong acid = Salt of strong acid + weak acid 

See Table A20 for list of weak acids. Some weak acids are unstable and 
decompose as formed. See tests 22-1, 22-9, 22-12, and 22-15. 


*(5) Salt of weak base + strong base = Salt of strong base + weak base 

See Table A21 for list of weak bases. 

*(6) Metal oxide + acid = Salt + H 2 O 

*(7) Metal sulfide + nonoxidizing acid = Salt + H 2 S 

Some sulfides with very low solubility react slowly or not at all; see 
Table A22. 

Balancing Redox Equations 

Reactions considered in this section are more complicated types in 
which oxidation and reduction (redox) take place. Oxidation of an 
element is defined as a real or apparent loss of electrons) resulting in a 
change from a lower to a higher (more positive) oxidation state. Reduction 
is the real or apparent gain in electrons) resulting in a change from a 
higher to a lower (less positive) oxidation state. Equations in which 
redox occurs are balanced by use of coefficients to equalize the electron 
loss and gain and simultaneously give an atomic balance. 

Oxidation numbers must be learned for some elements and others can 
be deduced as needed. Elements that arc uncombined or are present as 
polyatomic molecules have zero oxidation states. Elements in periodic 
group JA are + 1 (H may be 1 as in NaH), and in IIA are +2. Halogens 
not in combination with oxygen are 1 . Oxygen is 2 except in peroxides 
where it is 1 . Aluminum is (III). With these in mind one can find the 
oxidation numbers of elements combined with them since the sum of the 
numbers with their signs algebraically added is zero for a given compound. 
Considering HNO 3 for example, H is +1, the oxygen total is (3)(-2) 
= 6, so N must have an oxidation number of (V). This does not mean, 
however, that N+ 5 is capable of independent existence but it is only a 
notational convenience to be used in relating actual or assigned electron 
transfer in equation balancing, as will be illustrated later. Oxidation 
states are indicated by Roman numerals as N v , Mn vl1 , etc., or as nitrogen 
(V), manganese (VII), etc. The actual charge an ion has due to electron 
loss or gain, as Na^, K+, Ca+ 2 , Mg' 2 , Sr+ 2 , Ba^ Q-, Br~, and I~, is the 
valence or valence number. Since few other cases are so definitely electro- 
valent as these, the more general term oxidation state (or number) will 
often be used along with Roman numeral notation. See also Chapter 3. 
Of the metals to be analyzed in the laboratory the following exist in solution 
in combined forms other than simple hydrated ions and Roman numerals 
are used for their oxidation states: Bi m , As 111 , Sb m , Sn TI , and Sn iv . 
Trivalent ions like Cr f3 , Fe+ 3 , and Al' 3 are also frequently found in 


complexes (Chapter 4), and where there is doubt, the Roman symbols 
can be used. 





5 4 

I i 

3 2 

I i 

Oxidation >* 
(Electron loss) 


i i i i 

3 4 




1 1 


1 1 1 1 
f Reduction 
(Electron gain) 

1 1 




FIG. 1-1. A representation of rcdox. 

The Change in Oxidation State Method. Redox equations can be 
balanced merely by first equating the changes in oxidation number (electron 
transfer) and then balancing all other terms. The method involves these 
steps: (a) write the correct formulas for all substances, (b) Determine 
which elements change oxidation states, and balance the changes with 
coefficients. The elements may be connected with lines to follow the 
process. If an element directly concerned with the redox has a subscript, 
make that number a coefficient on the other side of the equation (since at 
least that multiple will be needed there), and then balance the total 
oxidation number change, (c) After the electron change is balanced, 
balance all other elements starting with those that occur the least number 
of times. If ions are represented, the total charges on each side must also 

Example 1-4. Potassium dichromate in acid soln. will oxidize oxalic 
acid, and the products will be chromic sulfate, potassium sulfate, carbon dioxide 
and water. The molecular equation is 

+ 12 (6e'Ktiin = reduction) -M> 

I I 

K.Cr 2 7 + H 2 S0 4 + H 2 C 2 4 = Cr,(SO 4 ) 3 + K,SO 4 + 2CO 2 + H 2 O 

I ) 

-f-6 (2rt~loH.s oxidation) 4-8 

The elements changing oxidation numbers are chromium (being reduced) and 
carbon (being oxidized). Cr has a subscript 2 on each side so one is ready to 
balance the total change in 2 Cr's which is 6 electrons. C has the subscript 2 
on the left side and none on the right so 2 is made a coeflf. before CO 2 , since at 
least that many carbons will be needed there. The total change of 2 electrons 
lost by the 2 C's must now be balanced against the 6 electrons gained by the 
2 Cr's. This balance is made by deducing that 6 is the least common multiple 
of electron loss and gain, so both are made equal to 6 by multiplication of 
coeffs. as needed. These will be 1 for each end of the Cr bracket and 3 for the 
compds. at each end of the C bracket. CO 2 , which was multiplied previously 


by 2, is now multiplied by 3, giving CO 2 a final cocff. of 6. The oxidation 
number changes are now balanced : 

K 2 Cr 2 7 + H 2 S0 4 + 3H 2 C 2 O 4 = Cr 2 (SO 4 ) 3 + K 2 SO 4 + 6CO 2 + H 2 O 

Next, the other atoms are balanced. One notices that there are 2 FCs on 
each side and that they check already. Since Cr and K are balanced on the 
right side, then no further change in right side SO 4 is expected; a total of 4 
SO 4 's are needed from H 2 SO 4 on the left. This gives 8 H's from 4H 2 SO 4 and 
6 H's from 3H 2 C 2 O 4 , enough to combine with the oxygen from K 2 Cr 2 O 7 to 
make 7H 2 O on the right side. Twelve O's from 3 H 2 C 2 O 4 's balance the O in 
6 CO 2 . The final molecular equation reads 

K 2 Cr 2 7 + 4H 2 S0 4 + 3H 2 C 2 O 4 = Cr,(SO 4 ) 3 + K 2 SO 4 + 6CO 2 + 7H 2 O 

Inspection of this shows that K+ and SO 4 2 were not altered in any way during 
reaction. Of the remaining substances, dichromate, hydrogen, and chromic 
ions will exist as such in soln., while oxalic acid (a weak acid), carbon dioxide, 
and water will remain largely in the molecular form. The correct ionic equation 
consequently is 

Cr 2 O 7 a + 8H^ + 3H 2 C 2 4 = 2Cr+ 3 + 6CO 2 + 7H 2 O 

Had this equation (without coeflf.) been originally given for balancing, the process 
would have been the same. As a final check one sees that the ionic charge is 
f 6 on both sides. 

Example 1-5. Balance: 

HN0 3 + CdS = Cd(N0 3 ) 2 -f NO + H 2 O + S 

The oxidation state of N in NO :l is (V), so, in going to NO in which it is (II), 
3e~ were gained. Nitric acid is therefore the oxidizing agent and N v is red. 
The oxidation number of sulfidc sulfur is 2 and it is oxidized to zero as free 
sulfur, so sulfide is the red. agent and electron donor, and its electron change 
is 2. By multiplying the electron gain by 2 and the loss by 3, a balance of 6 
electrons for each is obtained giving 

2HN0 3 + 3CdS = Cd(N0 3 ) 2 + 2NO + H 2 O + 3S 

Now a variation from the previous example is noticed. With 3 Cd's on the 
left, the correct coeff. for cadmium nitrate is 3, meaning 6 more NO^'s are 
needed than provided for when only the red. of N was taken into account. 
These 6 must come from HNO 3 , so 6 are added to the 2 already shown, account- 
ing for 2 N's, which change oxidation number and 6 that do not. The 8 H's 
now available will produce 4 H 2 O's and a check on O balance shows 24 on both 

8HNO 3 + 3CdS = 3Cd(NO 3 ) a + 2NO + 4H 2 O -f 3S 

The Half-Cell Method, (a) This method is explained in Chapter 11 
along with the use of Appendix Tables A 18 and A 19 which list reactions 



of oxidizing and reducing agents in acidic and basic solution. For pur- 
poses here, Tables 1-2 and 1-3 (which are adaptations of the more 
advanced information) may be used. In general, any substance listed 


(In Order of Decreasing Oxidizing Strength) 




Weight of 

Reduction Products 

F z + 2H+ 





F 2 





O 3 + 2H 1 



2 + H 2 

H 2 O 2 + 2H+ 




2H 2 O 

NaBiO :t + 6H f 




Bi 1 ^ 1 4- 3H..O + 

Na h 

KMnO! + 8H+ 




Mn-' 2 + 4H 2 O -\ 

- K+ 

PbO 2 + 4H f 




Pb |2 + 2H,,O 

CI 2 





K 2 Cr 2 O 7 + 14H 1 




2Cr 3 + 7H 2 OH 

h 2K+ 

MnO 2 + 4H+ 




Mn + 2H 2 

2 + 4H+ 




2H 2 O 

KCIO 3 + 3H+ 




HC1O 2 + H 2 O H 

\f. I* 

Br 2 





HN0 3 + 3H 1 (dil.) 




NO H- 2H 2 O 

Fe jn ' 




Fc 12 






among the oxidizing agents will oxidize any substance among the 
reducing agents. 

Example 1-6. Balance the reaction equation in which bismuthate ion 
reacts with ferrous ion in acid soln. 
From Tables 1-2 and 1-3, the reactants and products are found to be 

Bi0 3 + H+ + Fe+ 2 = Fe> 3 + H 2 O + Bi 111 

By further noting that the electron change in iron is one and in bismuth two, 
the final equation is 


2Fe+ 2 = 2Fe+ 3 + 3H 2 O 

When a half reaction in question cannot be found in tables, as might be 
the case in research work, predictions of reaction products can be difficult. 
Then analogies from periodic table relationships and a knowledge of other 
reactions of the particular element, its oxidization- states, and their 
interconvertibility are some guide posts in prognostications. 



(b) As a variation of the method, a given reaction can be divided into 
two half reactions, and, with the addition of H 2 O, H^ , or OH" (depending 
on acid or base solutions) to one side or the other of the half reactions, 


(In Order of Decreasing Reducing Strength) 

Reducing Mixture 



Weight of 

Oxidation Products 

Alkali metals 





Alkaline earth metals 




M+ 2 





A1+ 3 





Zn+ 2 





As + 3H+ 

H 2 C 2 4 




2C0 2 + 2H+ 





Fe+ 2 





Cd+ 2 

H 3 P0 3 + H 2 




H 3 PO 4 + 2H+ 






H 2 S 




2H+ + S 

Sn 11 




Sn lv 





Cu* 2 





Cu+ 2 





I 2 





Fc f3 

and adjustment of charge with electron additions, one is able to balance 
quite complex equations. 

Example 1-7. Balance, as described above: 
As 2 S a + HNO, = H : ,AsO 4 + N 

H 2 O 

This looks formidable because both As and S are oxidized and N is reduced. 
Dividing into two halves, we get 

As 2 S 3 = H 3 AsO 4 + S 
HNO 3 = NO 2 + H 2 O 

Division is always made so that one partial or half-reaction contains the oxida- 
tion and the other the reduction. The arsenic partial requires H 2 O on the left 
to supply hydrogen and oxygen. This balances with respect to atoms as 

8H 2 + As 2 S 3 = 2H 3 As0 4 + 3S + 10H+ 


The nitric acid partial requires H f on the left : 

H+ + HNO 3 = N0 2 4- H,0 

For balance of charge, the As half reaction needs 10^- on the right side, and 
the HNO 3 half reaction needs \e on the left. The two partials are balanced 
with each other by multiplying the HNO 3 reaction by 10, thus attaining over-all 
electronic balance. The whole reaction is obtained by adding the half reactions 
and canceling l(te , 10H f , and 8H 2 O from both sides: 

8H 2 O 4- AsaS 3 = 2H 3 AsO 4 + 3S 4- 10H+ 4- 10t>- 

IQe- 4- 10HNO 3 + 10H+ = \QNO 2 4- 10H,O 

As 2 S 3 4- 10HNO 3 = 2H 3 AsO 4 + 3S 4- 107VO> + 2H 2 O 

Note in this variation we were not concerned with oxidation states. 


1. Name the following substances. Refer to Tables A24 and A25 for nomen- 
clature aid: KC1O 3 , NaCIO, Ca(ClO 4 ) 2 , Mg(ClO 2 ) 2 , Bi(NO 3 ) 3 , Mnl 2 , CrF 3 , 
Pb 3 (PO 4 ) 2 , LiHCO 3 , Ti0 2 , CuSO 4 -5H 2 O, Mg(BO 2 ) 2 , SnCO 3 , Ni(OH) a , CuBr, 
A1 2 3 , Fe(C 2 H 3 2 ) 3 , AgCN, Hg 2 Cl 2 , Fe(CNS) 2 , Co(NO 2 ) 2 , CdS, As 2 (SO 3 ) 3 , 
Sb 2 (C 2 4 ) 3 , Sn(HS) 4 , BaCrO 4 , PbCr 2 O 7 , Hg(HSO 3 ) 2 , Zn 2 P 2 O 7 , (NH 4 ) 3 PO 3 , 
Na 2 B 4 O 7 -10H 2 O, (BiO)BiO 3 , SF 6 , CaMnO 4 . 

2. Give formulas for: silver iodide, lead (plumbous) chlorate, mercurous 
arsenate, mercuric (ortho)phosphate, bismuth metaphosphate, cadmium tartrate, 
cuprous sulfide, arsenous oxide, antimonic fluoride, stannous bromide, cobaltous 
cyanate, nickelic oxide, ferric thiosulfate, zinc metastannate, chromic carbonate, 
manganous bisulfate, aluminum bromate, calcium formate, magnesium sulfite, 
strontium hypochlorite, ammonium acetate, sodium peroxydisulfate, potassium 
permanganate, lithium hyponitrite. 

3. Write balanced equations showing formation of the following acids, using 
part I, C(3) of the type reactions compilation: H 2 CO 3 , HNO 2 , H 2 SO 4 , H 4 P 2 O 7 , 
H 3 PO 4 , HNO 3 , H 2 SO 3 , HC1O 2 , HC1O 3 , HC1O 4 . Match the names to the acids: 
chloric, sulfurous, sulfuric, chlorous, nitric, pyrophosphoric, nitrous, perchloric, 
carbonic, orthophosphoric. 

4. Predict the products (include indication of ions, molecules, gases, and 

solids) and balance: (a) CaO + H 2 O = ? (b) Mg + O 2 = ? (c) Mg + N 2 
= ? (d) Na + H 2 O = ? (e) A1(OH) 3 + HBr = ? ( f) potassium carbonate 
+ sulfuric acid = ? (^) sulfite ion 4- hydrogen ion = ? (/?) copper metal 
4- chlorine = ? (/) barium nitrite H- oxygen = ? (/) sodium carbonate 
decahydrate is heated = ? (k} CoO + C = ? (/) Li 4- hydrogen = ? (m) 
SDdium ferrocyanide 4- cupric nitrate = ? (n) ammonium chloride 4- barium 
hydroxide = ? (o) hydriodic acid 4- silver nitrate = ? (p) ferric acetate 


-f sodium orthophosphate = ? fy) ethyl alcohol is burned = ? (r) zinc 
sulfide -h HC1 = ? (s) aluminium + bismuth chloride = ? (/) bromine + 
Nad = ? (u) zinc + potassium hydroxide = ? (v) arsenic + magnesium 
sulfate = ? (w) mercuric oxide is heated = ? (x) manganese 4- acetic acid 
= ? (y) sulfuric acid -f aluminium cyanide = ? (z) lead -f AuCl = ? 

5. Predict the products and balance: (a) C1 2 + H 2 S = ? (h) Sn 11 -f H+ + 
PbO 2 = ? (c) KC1O 3 + H- + Cu+ = ? (d) KMn0 4 + H+ + Fe = ? (e) 
dichromate ion + hydrogen sulfide in acid solution = ? (f)l~ + H 2 O 2 + H+ 
= ? (g) ozone + arsine in acid solution = ? (h) BiO^ -h H 2 C 2 O 4 -f H+ - ? 
(/) manganese dioxide + stannous ion in acid solution = ? (y) H 3 PO ;J + per- 
manganate in acid solution = ? 

6. Indicate insoluble solids and gases, and balance by the change of oxidation 
state number method: 

(a) KI -f PbCrO 4 + HCI = PbCl 2 + KC1 + CrCl, -f I 2 + H 2 O 

(b) KMnO 4 + HNO 2 + H 2 SO 4 = MnSO 4 + K 2 SO 4 + H 2 O + HNO 3 
(r) Mn(OH) 3 + HCI + H 2 O 2 = MnCl 2 + H 2 O + O 2 

(d) KI -f NaBr0 3 -h HCI = KC1 + NaBr + H 2 O + I 2 

(e) Na 2 C 2 4 -f NaMnO 4 + H 2 SO 4 = Na 2 SO 4 + H 2 O + CO 2 -f MnSOj 
(/) MnSO 4 + PbO 2 + HNO 3 = HMnO 4 + PbSO 4 -f Pb(NO 3 ) 2 + H 2 O 
(g) NaBi0 3 + H 2 S0 4 -f NaAsO 2 = H 3 AsO 4 + H 2 O -f- Na 2 SO 4 + 812(804)., 
(h) HN0 3 + Bi 2 S 3 = Bi(N0 3 ) 3 + H 2 O + NO + S 

(/) CuS0 4 + KI = K 2 S0 4 -f Cul + I 2 

(/) CrI 3 + KOH + CI 2 = K 2 Cr0 4 + KIO 4 + KC1 + H 2 O 

7. Divide each of the reactions of problem 6 into two half reactions con- 
taining only the essential ions and molecules and balance them using the half 
cell method. 

8. Show reaction between bases of Li 1 , Mg+ 2 , and Al f3 and acids of Br~, 
NOijf, SO^ 2 , and PO 4 3 . Name each salt formed. 

9. (a) In problem 6, point out which substance is the oxidizing agent and 
which the reducing agent for each reaction, (h) Name each substance in each 

10. Reactions tend toward completion in the direction in which a precipitate 
forms. Which of these compounds are essentially water-insoluble: NaOH, 
HN0 3 , Cr(OH) 3 , Agl, Cd(NO 3 ) 2 , SnS 2 , CaC 2 O 4 , BaCrO 4 , K 4 Fe(CN) 6 , MgCl 2 , 
MnSO,, CaCO 3 , HgCl 2 , CuSO 4 . 

11. From the table of salt solubilities (back cover) one may draw the general- 
ization that almost all Na+, K+ and NHJ compounds are water soluble as are 
most salts containing C 2 H 3 O.J, ClOg-, NO 2 and NO 3 . What generalizations 
can one make on the water solubilities of (a) chlorides, bromides, iodides 
(b) phosphates (c) carbonates (d) sulfides, and (e) sulfates? 

12. Complete and balance the following half reactions, including electrons: 

(a) e~ -f CeO 2 + H+ = Ce+ 3 

(b) CuS = Cu+ 2 -h SO," 2 + 8H+ 

(c) Na = Na+ 


(d) 5e~ + MnCV = H 2 O + Mir* 8 

(e) I0 : r = H :{ 10 2 + 2e- 
(/) Br- + 6OH- = BrOjT 

( t ?) AsO 2 ~ + 2H,,0 = As + OH- 

(/) BiOCl + H* = Bi + Cl- 

(/) I 2 + H 2 = I0 3 + H* + i' 

1 3. What is the oxidation state of each element : HNO 3 , HMnO iv NaAl(OH) 4 
SbH 3 , Hg 2 Cr 2 O 7 , Hg(NH 2 )CI, Na a O 2 , BaCO 3 , KCr(OH),, 
14. How does one interpret the following: 

(a) Mn0 4/ , M + H. 2 S + Cl~ - r/ 2/ . r + S f + H a O + Mn+ 2 

(b) H ' + OH- r-> H 2 O + //, 

(c) Cut + Cd* 2 -f 8NH :J = 

15. Jackson P. Slipshod is a student capable of forming a snap opinion on 
any subject. He glances at one reaction equation and announces the Slipshod 
rule: "The number of compounds in a given molecular rcdox equation is the 
same as the number of elements involved." Inspect several more examples 
than J.P.S. inspected and modify his rule if you reach a different conclusion. 
How does the rule help balance equations when only the reactants are given? 


1. C. A. Vanderwerf, A. W. Davidson, and H. H. Sisler, J. Chcm. Ethtc., 22, 450 
(1945). (Redox) 

2. A. Standen,./. Chcm. Eihic., 22, 461 (1945). (Simple balancing) 

3. R. E. Oesper, J, Chem. Edut., 22, 290 (1945). (Modern nomenclature) 

4. A. Porges, J. Chem. Et/uc., 22, 266 (1945). (Balancing) 

5. G. Nakeham, J. Chem. Educ.< 23, 43 (1946). (Teaching valence) 

6. S. Glasstone, J. Chem. Etluc., 25, 278 (1948). (Oxidation numbers and valence) 

7. M. Martinette, J. Chem. Ethic., 26, 101 (1949). (Subcompounds) 

8. H. C. Brown and C. L. Rulfs, /. Chem. Ettoc.. 27, 437 (1950). (Problems in 
inorganic chemistry. Sec articles following also) 

9. L. P. Eblin,7. Chem. Educ., 28, 221 (1951). (Oxidation numbers) 

10. G. W. Bennett, J. Chem. Eiluc., 31, 324 (1954). (Material balance and redox) 





A solution is a homogeneous mixture whose composition can be varied 
and whose components are not easily intercon verted. The component in 
largest amount is the solvent, the other components are solutes. Solutions 
are classified in various ways: as dilute (ratio of solute(s) to solvent is 
small), concentrated (ratio of solute(s) to solvent is larger), or saturated (a 
maximum amouni of solute is in solution in equilibrium with excess solid 
solute at a specified temperature). Another classification is according to 
the original physical states of solvent and solute. Of the nine possibilities, 
three will be illustrated many times in the laboratory: a gas dissolved in a 
liquid (H 2 S in H 2 O), a liquid in a liquid (H 2 SO 4 in H 2 O), and a solid in a 
liquid (AgNO 3 in H 2 O). Water will be understood as the solvent unless 
otherwise stated. The solving of problems dealing with solutions depends 
upon exact expression of solution concentration, and the following 
sections explain the most often used terms and methods. 

Weight Ratio 

Some handbook solubility data are given in grams of solute per 100 g of 
solvent. Determination of solubility and its expression are illustrated. 

Example 2-1. 20.00 ml of satd. KC1O 3 soln. at 20 C weighs 21.000 g, and 
after low temp, vacuum evap. of the water (to prevent decompn. of KC1O 3 by 
heating), the residue of KC1O 3 weighs 1.450g. Calculate the soly. in g/lOOg 
H 2 0. 



The data shows that the wt ratio of solute/solvent at 20 C is 1.450/(2 1.000 - 
1.450). Solving for that wt of solute which is equiv. to 100 g of solvent gives 

? g KC1O 3 = 100 g H 2 (1.450 g KCIO 3 / 19.550 g H 2 O) 
= 7.425 g KC10 3 / 1 00 g H 2 O. (Ans. ) 

Solution concentration in these terms is generally reserved for the discus- 
sion of saturated solutions and is an expression of the maximum equilibrium 

Weight Percent 

The weight per cent of a solution refers to per cent solute and implies 
that when the solution is to be used, its weight and not its volume will be 

Example 2-2. Calculate the wt % KC1O : , in the soln. of Example 2-1. 
The calcn. infers the statement: 

Wt % = (wt solute/wt solute + wt solvent) 100 < / , ((2-1) 

= 100 (wt solute/wt solution) 

Substitution gives 6.90 wt % KC1O 3 . (Ans.) 

The student should observe in these two examples that no molecular 
weights entered the definitions or calculations. 

Weight ratio and weight per cent arc two methods of solution concen- 
tration expression inphvsical units. The methods following are in chemical 

Formality (F) 

The formality of a solution is the number of gram formula weights 
(form, wt) of solute dissolved in a liter of the solution: 

F = g solute per liter of solution/formula wt solute (2-2) 

= Formula wts of solute per liter of solution 

Thus a measured volume of solution contains a known weight of solute. 

Molarity (M) 

The molarity of a solution is the number of gram molecular weights 
(moles) of solute dissolved in a liter of the solution: 

M = g solute per liter of solution/mol. wt solute (2-3) 

= Moles of solute per liter of solution 


Example 2-3. One dissolves 39.215 g of Mohr's salt, FeSO 4 -(NH 4 ) 2 SO 4 - 
6H 2 O (392.15)* in enough H 2 O to make 10 liters of soln. Calculate F and M 
with respect to the solute and M with respect to each ion. 

The soln. contains 3.9215 g of solute per liter, and since the mol. wt and 
form, wt are identical, then with regard to the entire solute, 

F = M = 3.9215 g/392.15 = 0.0100 (Am.) 

The compd., as a typical salt, will be considered to dissoc. completely, each 
molecule giving Fe+ 2 + 2SO^ 2 + 2NH 4 . Therefore with regard to the ions, 
the soln. is 0.0100 M in Fe+ 2 , 0.0200 M in SO 4 '-, and 0.0200 M in NH 4 . (Ans.) 

Because ionic compounds are composed of ions both in the solid state 
and in solution, the term formality rather than molarity is generally used 
to describe concentrations with respect to the entire solute. A square 
bracket around the abbreviation for a substance like [Fe 12 ] signifies a 
molar concentration. Molarities will be used more often in this text. 

For small quantities of solutions, mg instead of g, and ml instead of 
1 (liter) arc employed. One then defines a millimole (mmole) as a milli- 
gram molecular weight, and a molar solution will contain a mg mol. wt per 
ml of solution. 

Example 24. A soln. is 1.50 M in FcSO 4 . Calculate the mg and mmole 
of Fe+ 2 in 1 ml. 

If the soln. is 1.50 M in FeSO 4 , it is 1.50 M in Fe 1 2 , and the ionic wt of Fe^ 2 
(55.85) can be used in place of mol. wt in equation 2-3): 

[Fc+ 2 ](55.85) = g Fe 2 per liter of soln. = 83.8 g Fe+ 2 per liter 

= 83.8 mg Fe+ 2 per ml (Ans.) 

Since [Fe 42 ] = 1.50 M, there are 1.50 moles of Fe 42 per liter = 1.50mmoles 
Fe+ 2 per ml. 

Molality (m) 

The molality of a solution is a measure of the number of moles of 
solute dissolved in a kilogram (lOOOg) of solvent. This concentration 
expression is also known as weight formality and weight molarity. Solu- 
tions concerned with boiling point, free/ing point, and osmotic pressure 
changes are made up using this concentration system. Using a given 
solvent, one notes the interesting feature that all solutions of the same 
molality have the same ratio of solute molecules to solvent molecules. 
One weighs the solution to be used instead of measuring its volume since 
when one dissolves a weight of solute in a weight of solvent, there is no 
way to theoretically calculate the final volume, as the volumes (except 

* Appendix tables give mol. wts for use in problem solving. Slide rule accuracy is 


those of dilute aqueous solutions) cannot be assumed additive. The final 
volume may be experimentally observed for an individual mixture but 
these are not tabulated in reference books. 

m = g solute per 1000 g solvent/mol. wt solute (2-4) 

= Moles of solute per kg of solvent 

Example 2-5. Calculate the molality of the soln. from Example 2-1. 
As in that example, one first determines that the soln. contains 74.25 g KC1O 3 
(mol. wt 122.5) per 1000 g of H 2 O. The molality from equation 2-4 is 

iw = 74.25/122.5 = 0.606 (Ans.) 

Mole Fraction (n) and Mole Per Cent 

The mole fraction (n) of a solution component is the number of moles of 
that component in the solution divided by the total moles of solution 
components present. Mole per cent is 100/7. The sum of all mole per 
cents for a given solution will be 100% and of all mole fractions will be 
1.00. Problems concerning Raoult's laws, for example, involve mole 
fraction computations. For component A in a mixture of A + B + . . . Z, 

n v = Moles A /(moles A + moles B + moles Z) 

Example 2-6. A soln. is made by mixing 100 g of cthylene glycol (HOCH 2 
CH 2 OH = 60.0) and lOOg of H 2 O (18.0). Calculate the mole fraction of 
each component. 

Moles H 2 O = 100/18.0 = 5.55 

Moles glycol = 100/60.0 = 1.67 

Total moles = 5.55 + 1.67 = 7.22 

WjT20 = moles H 2 O/moles soln. = 5.55/7.22 = 0.769 (Ans.) 

glycol moles glycol/moles soln. = 1.67/7.22 = 0.231 (Ans.) 

"H/> + "iibroi = 0-769 + 0.231 = 1.000 (Check) 

Normality (N) 

The normality of a solution is an expression of the number of gram 
equivalent weights of solute present in a liter of solution. The gram 
equivalent weight is also called a gram equivalent or just an equivalent. 

N = g solute per liter of solution/equiv. wt solute (2-5) 

= Equivalent of solute per liter of solution 

Although this system is very useful in calculations for analytical opera- 
tions it needs careful definitions, particularly of the equivalent weight. 
The following brief treatment will be sufficient for our purposes. It is 
given in more detail in quantitative analysis courses. 


A gram equivalent weight of a substance is that gram weight which will 
react with or is otherwise related to 8.0000 g of oxygen. For instance, in 
a mole of H 2 6 weighing 18.0160 g, 2.0160 g of H are combined with 
1 6.0000 g of O, or, by the preceding definition, the gram equivalent 
weight of H is l.OOSOg. Since this is related to 8.0000 g of O, it may 
itself be used as an equivalent weight standard; for instance Cl in HC1 is 
found then to have an equivalent weight of 35.457, etc. For substances 
not entering into redox reactions (considered later in this chapter), the 
equivalence system is set up on a univalent basis. Though needing later 
modification, a useful starting point is 

Equiv. wt = Form, wt/valence number (2-6) 

It follows that equal volumes of solutions of like normalities will be 
chemically related to each other, since they all contain the same number 
of equivalent weights of solute. 

Three cases not concerned with redox are important: 

(1) Equivalent weights of salts are found by dividing the formula 
weight by the total valence number (regardless of sign) of the ion for 
which normality is to be expressed. 

(2) The equivalent weight of an acid is that weight which furnishes 
1.0080 g of H+ (or reacts with 17.0080 g of OH~) in a specific reaction. 

(3) The equivalent weight oj a base is that weight which furnishes 
1 7.0080 g of OH (or reacts with 1 .0080 g of H ) in a specific reaction. 

Thus, for example, Al 2 (SO 4 ) 3 /6 is the equivalent weight of this salt, 
H 3 PO 4 /2 is the equivalent weight of this acid if used as H 3 PO 4 + 2NaOH 
= Na 2 HPO 4 + 2H 2 O, and Ca(OH) 2 /2 is the equivalent weight of this base. 

Example 2-7. A bottle of coned. H 2 SO 4 has this data on the label : "Sp. gr. 
25/4 = 1.840, Assay = 96? H 2 SO 4 by wt, Form, wt = 98.08." 

Find Fand N for this soln. assuming both H' will be used in reaction. 

The sp. gr. term in the data means that the density of H 2 SO 4 at 25 C is com- 
pared to the density of H 2 O at 4 C, and since the latter is 1 .000 g/cc, then 
1.840 g/cc is the density of H 2 SO 4 at 25 C.* One can therefore find the wt of 
H 2 SO 4 per liter by multiplying the wt per cc by 1000 and multiplying that by 
0.96. This establishes the numerator for finding both F and N, using equations 
2-2 and 2-5: 

F = ( 1 .840 g/cc)( 1 000 cc/liter)(0.96)/98.08 = 1 8.0 (Ans.) 

N=(\ .840 g/cc)( 1000 cc/liter)(0.96)/(98.08/2) = 36.0 (Ans.) 

* Sp. gr. 25/25 would mean the density of the substance in question at 25 C is com- 
pared to the density of H 2 O at 25 C (0.9971 g/cc). This is not numerically equal to 
the density of the substance at 25 C, though in this problem and this course the distinc- 
tion is not important. Sp. gr. 25/25 is the usual type measured in the lab. 

The milliliter and cubic centimeter will be treated as equal volumes. For accurate 
calculations, one multiplies the number of ml by 1.000028 to convert to cc. 


Normality, Molarity, and Formality Relationships 

As anticipated in Example 2-7 a simple relationship exists between 
M, N, and Ffor a given solution. It is seen by comparison of formulas 
2-2, 2-3, and 2-5 that the terms are related as molecular, equivalent, 
and formula weights are related. For a given solution component, N 
will either be the same as M (or F), if mol. wt = equiv. wt, or larger by 
a multiplying factor that has the- value, mol. wt/equiv. wt. In Example 
2-7, since the mol. wt or form, wt is twice the equiv. wt, normality could 
have been found by N = (98.08/49.04)(F). 

Working with Equivalents and Milliequivalents 

A liter of one normal solution contains one equivalent of reactive 
component, a half liter of this solution would contain half an equivalent, 
a tenth liter of half normal solution would contain (0.1)(0.5) = 0.05 
equivalents of solute, etc. In other words, for solutions, 

(litarXN) = Equivalents (2-7) 

and if a milliequivalent (meq.) is defined as 0.001 equivalents, then 

(m\)(N) = Milliequivalents (2-8) 

The terms can be related to weights of reacting materials as well as to 
solution volumes. Considering H 2 SO 4 as in Example 2-9, since 49.00 
is the equivalent weight, then 49.00 g is one equivalent, 98.00 g = 2 equi- 
valents, etc., or for weights of pure reactants 

g equiv. wt == Equivalents (2-9) 

and as before 

mg/equiv. wt = Milliequivalents (2-10) 

These mathematical definitions of the equivalent and milliequivalent 
are useful in problem working. If A and B are going to react or be 
chemically related, then the first statement one may make is, equiv. A 
= equiv. B or meq. A = meq. B. 

Example 2-8. How can one prep. 200 ml of 0.50 N H 2 SO 4 from 1.22 M 
H 2 S0 4 ? 

The vol. of more coned, acid used for diln. and the vol. of dild. acid obtained 
must contain the same number of meq. of H 2 SO 4 , or meq. coned, acid = meq. 
dil. acid. In the general case, 

(Ml coned. soln.)(yV coned, soln.) - (ml dil. soln.)(/V dil. soln.) (2-11) 

The more coned, acid is 1.22 M or 2.44 N. (Why?) Solving equation 2-11 
gives 41.0ml. This vol. of 1.22 M acid is dil. with water to a total vol. of 
200ml. (Ans.) 


Example 2-9. fen ml of a sodium sulfate soln. having a density of 1 . 10 g/cc 
25/4 arc mixed with excess barium chloride soln. and the resulting ppt. of 
barium sulfate weighs l.OOOg. Find the normality of the sulfate soln. and 
% by wt Na 2 SO 4 in it. 

In the reaction, meq. Na 2 SO 4 = meq. BaSO 4 . Since Na 2 SO 4 is a soln., meq. 
Na 2 SO 4 = (ml)(AO, and for the solid, meq. BaSO 4 = mg/equiv. wt. The equiv. 
wt of BaSO 4 is half the mol. wt (by equation 2-6), and equating the two defini- 
tions for meq. one gets (ml Na 2 SO 4 )(W Na 2 SO 4 ) = mg BaSO 4 /equiv. wt BaSO,. 
Solving for soln. normality gives N = 1000 mg/(10.0 ml)(233/2) = 0.859. (Ans.) 

The meq. of Na 2 SO 4 are now found either by 1000 mg BaSO 4 /(233/2) or by 
(10ml Na 2 SO 4 )(0.860). Either gives 8.60 meq. From equation 2-10, the mg 
of Na 2 SO 4 can be determined since 

mg = (Meq.)(cquiv. wt) = (8.60)(142/2) = 610 

The % solute in the soln. will be found according to equation 2-1, the soln. 
density being used to find the soln. wt: 

% Na 2 S0 4 = (610 mg Na a SO 4 /l 1,000 mg soln.) 100% = 5.54% (Ans.) 

Problems like the last one and those in which a substance is determined 
by titration of a weighed sample are characteristic of quantitative analysis. 
A general formula one can deduce from the stepwise method above for 
find the % by weight of a rcactant AT in a mixture is 

% X = (meq. X)(equiv. wt X)(100%)/mg sample (2-12) 

Normality and Redox 

Equivalent weights of substances undergoing oxidation state changes 
are calculated in a different way from examples cited thus far. It can 
be experimentally shown for example that in dilute H 2 SO 4 solution, to 
oxidize 55.85 g of Fe f2 to Fe+ 3 , requires 1/5 mole of KMnO 4 , 1/6 mole of 
K 2 Cr 2 O 7 , or 1 mole of Ce(SO,) 2 . These quantities of oxidizing agents 
must therefore be equivalent to one another, and examination of their 
oxidizing reactions 

5e~ + MnOj + 8H+ = Mn+ 2 + 4H 2 O 
6<>- + Cr 2 0- 2 + 14H*- = 2Cr+ 3 + 7H 2 O 
e - + Ce+ 4 = Ce+ 3 

shows that the comparison is on the basis of the number of electrons 
gained per molecule or ion, or their change in oxidation number. Re- 
ducing agents are related by the number of electrons donated per molecule 
or ion, and their corresponding change in oxidation number. The normal 


system for redox solution concentrations is therefore constituted on a 
unit change in oxidation number. 

Equiv. wt redox agent = Form, wt /e" change per molecule or ion (12-13) 

If one considers a gram mole of reactant, then the equivalent weight is 
that gram weight which will accept or donate 6.02 x JO 23 electrons. The 
electron change is found in tables of redox half reactions (Tables 1-2, 
1-3, A 18, and A 19) or found when the equation is balanced. Other 
considerations for problem solving are identical with those already 

Example 2-10. One dissolves 25.00 g of K 2 Cr 2 O 7 (form, wt 294.2) to make 
1 liter of soln., and 45.5 ml of this will just oxidize the iron (Fe f2 = Fe+ 3 ) in 
1780mg of properly dissolved iron ore. Calculate N of the oxidizing agent, 
equiv. wt of Fe, and % Fe in the ore. 

As noted above and in Tables 1-2 and A 1 8, the change in oxidation number 
for dichromate is 6, so the equiv. wt is 294.2/6 = 49.0, and since the solution 
contains the 25.00 g of K 2 Cr 2 O 7 per liter, N = 25.00 g per litcr/49.0 = 0.510 TV 

Iron undergoes a one electron change per ion in the oxidation, Fe +2 = Fe+ 3 , 
so its ionic and equiv. wts are the same, 55.85/1. (Ans.) 

The mcq. of K 2 Cr 2 O 7 = (ml)(/V) = meq. of Fe+ 2 . The problem with this 
much elucidated then follows the type solution described by equation 2-12: 

/ Fe = (45.5 ml K 2 Cr 2 O 7 )(0.510 N K 2 Cr 2 O 7 )(55.8/l equiv. wt Fe+ 2 ) 
( 1 00 %)/] 780 mg sample 

Converting the figures to exponential numbers* for ease in approx. the answer 

= (4.55 x 10)(5.10 x 10 A )(5.59 x 10)(10 2 )/1.78 x 10 3 

The ten's cancel : 

= (4.55)(5.10)(5.59)/1.78 
This is 

and using the slide rule gives the correct value, 72.9% Fe. (Ans.) 


1. A solution is made by dissolving 40.0 g of Ca(NO 3 ) 2 -4H 2 O (form, wt 
236. 1 6) in 380 g of H 2 O giving a final volume of 400 ml. Calculate (a) F (b) M 

* See Appendix, mathematics review, Section Al and A3. 


in Ca+ 2 (c) M in NO 3 (d) m in Ca+ 2 (e) m in NO^ (f) density (g) N in Ca+ 2 
(A) AT in NO 3 - (/) % by wt solute. [Ans. (a) 0.423 (6) 0.423 (c) 0.846 (/) 0.447 
W 0.894 (f) 1.05 g/cc (#) 0.846 (h) 0.846 (/) 9.53%.] 

2. How can one prepare the following? 

(a) IOOmlof0.100A/Na 2 SO 4 

(b) 250 ml of 0.250 AT NaOH 

(c) 10.5 liters of 0.0022 N H 2 SO 4 (use as diprotic acid) 

(d) 55.0 ml of 0.10 N KMnO 4 (use as oxidizing agent in acid) 

(e) 100 ml of 0.20 N FeSO 4 (use as red. agent in acid) 

3. One wishes to oxidize 0.01 M H 2 S (acidified) solution. List these oxidants 
in decreasing order of their capacity: (d) 4.5 ml of 0.02 M KMnO 4 (h) 6.3 ml 
of 0.01 M K a Cr a 7 (c) 8.1 ml of 0.10 N Ce(SO 4 ) 2 (d) 25 mg of NaBiO, (e) 
0,020 g of MnO 2 . 

4. A solution contains lOmg Cu +2 per ml and was prepared by dissolving 
CuSO 4 - 5H 2 O (form, wt 249.7) in water. Find (a) g cupric sulfate pentahydrate 
in a liter of solution, (b) F (c) N (d) M with respect to each ion, and (e) the 
weight of BaSO 4 one could obtain by adding excess BaCl 2 to 10 ml of the 
copper solution. 

5. One dissolves 18 g H 2 C 2 O 4 to make 500 ml of soln. Calculate N based 
on these reactions: 

(a) H 2 C 2 O 4 + NaOH = NaHC 2 O 4 + H 2 O 

(b) H 2 C 2 4 -f 2KOH = K 2 C 2 4 + 2H 2 O 

(c) H 2 C 2 O 4 + Pb0 2 + H+ = H 2 O + CO 2 + Pb 2 (balance) 

(d) H 2 C 2 O 4 + Ca+ 2 = CaCjjOj + 2H f 

6. Express each in moles, mmoles, equivalents, and meq.: 

(a) 150 g 20% H 2 O 2 (oxidant in acid) 

(b) 40 mg NaCl 

(c) 2.79 g Sn 11 (reductant in acid) 

(d) 1 liter O 3 gas, C, 1 atm pressure (oxidant in acid) 

(e) lOmgZn. 

7. Thirty-five ml of an NaOH soln. will just neutralize 1.32g of KHSO 4 . 
Find (a) how many meq. of acid and base are used (b) reaction equation (c) mg 
of NaOH used (d) N of base, and (e) mg of H 2 O formed. 

8. From which of these can one make the largest volume of 0.1 N soln.? 
(a) 12.6 g H 3 PO 4 (as a triprotic acid), (b) 6.80 g Kl (as a reductant in acid), 
(c) 9.20 g KC1O 3 (as an oxidant in acid), (d) J3.3 g Ba(OH) 2 (as a dihydroxy 


9. A solution of H 3 PO 4 has a sp. gr. of 1.75 25/4 and is 90% phosphoric 
acid by weight. Find (a) F (b) N as a triprotic acid (c) volume needed to furnish 
l.OOg H+ (d) volume needed to make 1 liter of 0.20 TV acid (e) volume needed 
to make 280 ml of 0.14 F acid. 

10. Balance the equation: 

As 2 S 5 + HNO 3 = H 3 AsO 4 -f H 2 SO 4 -f ATO 2 + H 2 O 
(a) Find the equiv. wts of the redox agents. 



(b) If the HNO 3 used had a sp. gr. of 1.42 25/4, and was 72% acid by wt, 
find N. 

(c) What vol. of this acid would theoretically be needed to react with 50 mg 
of As 2 S 5 ? 

11. The solubility of two typical salts is given below in g/100 g H 2 O at various 
C temperatures, (a) Plot C on the abscissa and put both curves on the same 
plot, (b) What is the solubility of each at 50 C ? (c) If one cools a saturated 
solution of each from 90 C to 10 C, what weights of salts crystallize? (d) What 
is the % by wt solute of each solution at 30 C ? Can one determine the for- 
mality of either solution from this data alone ? If not, what other measurements 
would be needed? (f) At what temperature is their solubility the same? 
(g) What is the molality of each saturated solution at 25 C? 







K 2 Cr 2 7 (g/100 g H 2 0) 







NH 4 C1 (g/100 gH 2 0) 







(Ans. (b) 34.0 and 50.4 (c) 63.0 and 38.0 (d) 16.7 and 29.3 (/') 91 C (g) 0.51 
and 7.10.) 

12. How does one prepare: 

(a) A liter of KMnO 4 solution of such -normality that each ml would oxidize 
6 mg of Fe+ 2 ? (Ans. 3.38 g/liter = 0.107 N.) 

(b) A liter of HC1 solution of such normality that each ml would be equivalent 
to 1 % NaHCO :i in reacting with a 1.000-g sample of impure NaHCO a ? (Ans. 
4.35 g/liter = 0. 119 N.) 

13. Jackson Slipshod dissolves 50 g of AgNO : , in 50 ml of H 2 O giving 100 g 
of solution. He computes the normality as N = 500 g per liter/170 mol.wt 
= 2.94. Jackson's Uncle Frisbee (who helps Jack with home work) says that 
a weight of solution and not its volume was used in the calculation so what he 
actually found was solution molality. What goes on here? 

14. What weights of alum, K 2 SO 4 A1 2 (SO 4 ) 3 24H 2 O, (mol. wt 949) are needed 
to prepare solutions 1.0 N in K+, in SO 4 2 , and in A1+ 3 ? (Ans. 474 g, 1 19 g, 
158 g). 

15. One has a closed piping system for hot water circulation. One Ib of 
potassium dichromate is dissolved in the water and analysis after mixing shows 
the solution contains 5 mg of Cr VT per liter. Find (a) the solution F and (b) the 
gal of water in the pipes. 







The first modern era atomic model was proposed in about 1900 by 
J. J. Thompson. He visualized atoms as spheres containing plus and 
minus charges, respectively protons and electrons, scattered uniformly 
throughout. He supported this conclusion by calculating that in such 
a model electrons would vibrate approximately at the frequency of visible 
light, and hence be able to give the familiar flame tests. The Thompson 
model lasted until 1911 when Ernest Rutherford, Nobel laureate and the 
father of atomic physics, announced another concept of the atom which 
is a helpful visualization even today. His model consisted of a very dense 
nucleus containing positive protons, equal in number to the atomic 
number, surrounded by a cloud of negative electrons sufficient to make 
the atom neutral. He was able to show that the radius of the nucleus is 
in the order of 10~ 12 cm, whereas the radius of the whole atom including 
the electron shell is about 10 8 cm. The experiment that led him to these 
conclusions was the now classical one involving bombardment of a very 
thin gold foil with a-particles (helium nucleii from a radioactive source) 
and observation of the emergent beam as it traversed a Wilson cloud 
chamber behind the foil. This chamber is a device containing air super- 
saturated with water vapor in which moving charged particles ionize air 
molecules and cause water droplets to condense in their wake and render 
their paths visible. From the fact that the positive nucleii of atoms in 
the foil scattered so relatively few a-particles, Rutherford concluded that 



matter is essentially porous and that Thompson's model did not fit the 
experimental evidence. From the scattering angles and number of oc- 
particles deflected, Rutherford also computed the atomic and nuclear 
dimensions given above. 

In 1913, the Danish physicist Niels Bohr, who later received a Nobel 
Prize for his work, began studies to extend the atomic concepts to account 
for more of the known fundamental concepts of electromagnetism than 
did any of the atomic models of the time. The idea of a cloud of electrons 
somehow surrounding the nucleus of the atom did not appear satisfactory 
and consistent with his experience in related physical measurements for 
several reasons: 

(1) It is too vague about an orderly electron arrangement which one 
instinctively feels is present because all atoms of a given element have 
the same definite chemical properties. 

(2) If the electron vibrates in its orbit, and it does, it should give off 
some sort of radiation, but this would mean decreasing electron energy 
and a gradual spiraling of the electron into the nucleus. 

(3) If electrons are in the process of dropping toward the nucleus, they 
should be found at all distances from it, and gaseous elements should give 
a continuous spectrum corresponding to all energy states for electrons 
under the influence of an electric discharge. Such is not the actual case, 
however, because Balmer in 1885, Paschen in 1896, and others since 
showed that elements give spectra consisting of lines, which can be 
interpreted best if one assumes that electrons have a limited number of 
fixed energy states. 

Line Spectra and Energy Levels 

The experiments on the line spectra of hydrogen arc important since 
they yielded fundamental information on electrons. The experimenters 
found that no energy was emitted when hydrogen atoms were bombarded 
in an electric discharge tube by a beam of speeding electrons until the 
energy of the beam was increased to a certain value, whereupon the process 
of energy transfer suddenly became efficient. At this point many of the 
bombarding electrons lost their speed and the energized H atoms which 
had been struck gave off light of characteristic wavelength. By using 
increasingly energetic electrons it was found that there are only a limited 
number of energies that are absorbed and that result in a series of energy 
emissions from the excited atoms. These energies are manifested as 
spectral lines that can be sorted out with a spectrometer or spectroscope 
(special experiment 5). If the lines are in the visible region, they constitute 
an optical spectra. Excited molecules are also capable of electronic 
disturbances and of giving spectra. 




- X-rays - 

-Radio waves- 

1(T 2 1 

-1 1 

10 2 


10 4 


10 6 


10 8 


10 10 



, t 


- Ultraviolet - 

- Infrared - 

FIG. 3-1. Electromagnetic radiation. Wavelenths are in Angstrom units (1 A 
= 10 8 cm). See Fig. 23-2 for an expansion of the visible region. 

Optical spectra come from a disturbance of valence electrons, or, if 
greater excitation is involved there, ultraviolet spectra is the result. A 
particular type disturbance of the K and L electrons results in X-ray 
spectra. Infrared spectra are associated with internuclear disturbances of 
vibration and rotational motion. Measurements on these phenomena 
have proved invaluable in atomic and molecular structure studies. Sub- 
sequently all the elements were investigated in this way. All give line 

+j co 



5 ^ 



a 1 


rH g 

1 i 

-1 o 

s > 

1 1 

t ) 


/ ' 

Wavelength (X) >- 

FIG. 3-2. The Balmer series of spectral lines obtained by exciting H atoms. 

Up to 1913 no satisfactory correlation was made between the spectral 
data and the structure of atoms, though it seemed obvious that there 
should be a connection of fundamental significance. Then Bohr gave his 
interpretation based upon the quantum theory and the great puzzle of 
electronic structure began to be solved. In 1900 Max Planck, and in 
1905 Albert Einstein, had shown that certain radiation effects could be 
satisfactorily accounted for if one postulated that light is absorbed and 
emitted in definite units (called quanta) having energy hv* Bohr began 
with the assumption that the hydrogen spectra was due to energy emitted 
by the electron, which, having been knocked out of its normal position, 
was returning to that state of lower energy. It followed that the extra- 

* The symbol h is Planck's constant, 6.624 x 10~ 27 erg sec, and v is the frequency of 
radiation in cycles/second. It is defined as c/A where c is the speed of light, 3 
cm/sec and A is the Angstrom wavelength of the radiation being studied. 



nuclear electron did not have arbitrary energy but existed in definite energy 
levels called stationary states, and a definite amount of energy was needed 
to displace the electron from a lower energy level (closer to the nucleus) 
to a higher one (further from the nucleus). Conversely this unit of 
energy was released when the electron dropped back to its original position. 
If E! is the energy of a stationary state of higher energy and 2 the 
energy of the next state of lower energy, then a quantum of energy is the 
difference between them, since the electron can exist only in these discrete 
states and in no intermediate ones, or 

^ - 2 = hv 

v is the frequency of radiation needed to cause the electron to jump from 
E 2 to E l% or the frequency emitted when it falls from E l to E 2 . Pro- 
ceeding with this concept Bohr was able to calculate the wavelengths of 

15 r- 






FIG. 3-3. Energy level diagram for hydrogen. To remove the electron from its 
normal state, n = 1, to infinity, n cc , requires 13.60 electron volts (ev), its ionization 
potential. One ev is the energy gained by an electron when accelerated by a potential 
of one volt. One ev is equal to 1.59 x 10~ 12 ergs and corresponds to an energy of 

23.05 kcal/g mole. 


hydrogen spectral lines and the radii of electron energy levels from data 
and physical constants established by earlier investigators by using a 
variety of different experiments, some not directly related to hydrogen 
spectra. The concept of energy quantization paved the way for the 
determination of all electronic configurations and chemical bonding as 
well, two foundations of chemistry. 

When the energy levels of an atom are determined from its spectra, an 
energy level diagram can be prepared. One of the most significant 
values in it is the energy needed to remove the least tightly held electron 
from the atom to a distance sufficient so that the positive nucleus no 
longer exerts any attraction on it. This energy is called the fonization 
energy or potential. The energies associated with the removal of successive 
electrons can also be found experimentally. It is evident that chemical 
properties are related to these potentials since atoms form chemical 
bonds according to their abilities to effect a transfer or sharing of electrons. 
This is considered in a later section in the chapter (see Table 3-5). 

Refinement of the Bohr Atom; Quantum Numbers 

Study of spectra of elements with instruments of greater resolving 
power followed Bohr's work, and the new data showed a need for extending 
the Bohr theories. It was demonstrated, for example, that more spectral 
lines existed than were originally believed present. This meant that 
there were more subtle electronic energy changes than formerly proposed; 
as each of these may be quantized, more quantum numbers were needed 
to describe the electron. It was then found that certain rules or selection 
principles need be applied to the handling of the numbers for proper 
accounting of the observations. The reasons behind these will not be 
developed here, but the rules are summarized below. 

(1) The principal quantum number n can have integral values 1, 2, 
3 ... denoting three dimensional shells of energy in which the electron 
might be found. These are alternately called the K, L, M, N, O, P, and 
Q shells. The shell closest to the nucleus is described by n = 1 , the 
next n = 2, etc. If n is large, the electron cloud around the nucleus is 

(2) For each value of n there are n sublevels within a shell. These are 
differentiated by the quantum number /, known as the secondary or 
azimuthal quantum number. I is linked with the angular momentum of 
the electron. It may have values 0, 1, 2 ... (n 1). Electrons for 
which / = 0, 1, 2, 3 are respectively called ,?,/?, </, /electrons, after spectro- 
scopists' early nomenclature for spectral lines (sharp, principal, diffuse, 
/undamental). A Is electron is thus an electron in the first shell for which 
/ = 0; a 4f electron is in the fourth shell with / = 3, etc. 









First = K 



Second = L 

0, 1 

S 9 p 


Third = M 

0, 1,2 

>v, p. d 


Fourth = N 


s,p,d, f 

The / quantum numbers are important in chemical bonding considerations 
and arc considered again under covalcnce in this chapter. For our 
purposes, there is no / value greater than 3, so s, p, d, and /'electrons will 
be the only types. / gives the shape of the electron distribution: the 
simplest is spherical, the next is figurc-8 shaped, etc., as explained later. 
(3) In 1896, P. Zceiuan had noted that spectral lines that normally 
appear singly may appear as doublets under the influence of a magnetic 
field. To explain this, the quantum number m was introduced. This is 
related to a part of the total angular momentum of the electron in the 
direction of the magnetic field. The quantum theory shows mathemati- 
cally that if this is quantized, it can have all integral values between 
/ and / including zero, in other words a total of (21 + 1) values, m 
relates the orientation of the electron clouds to an externally applied 
magnetic field. For us, the magnetic quantum number will be of minor 








-1,0, 1 



-2, -1,0, 1,2 



-3, -2, -1,0, 


(4) Spectral lines under high resolution often appear as multiplets 
lying close together rather than as single, relatively broad lines. This 
was not explained until 1925 when G. Uhlenbeck and S. Goudsmit showed 
mathematically that the effect could be explained if the electron were con- 
sidered to be spinning, that the spin contributed to its total angular 
momentum, and that this contribution might be quantized. The spin 
quantum number is designated y and can have only the values \ and | 



representing opposite spins and corresponding to two levels of nearly 
equal energy. 

(5) W. Pauli announced a unifying principle in 1925 that makes it 
possible to properly assign energy states to the electrons in an atom. 
One might, for instance, logically suppose that since the first orbit is the 
one with lowest energy, electrons should be found only in that one. 
Pauli proposed, however, that it is impossible for two electrons in the 
same atom to have all four quantum numbers, w, /, w, and s, alike a 
statement known as the Pauli exclusion principle. This means that a shell 
may have only as many electrons as the quantum numbers and the Pauli 
principle permit. The total number of electrons permitted in a shell is 
found to be 2w 2 and the number of shells in normal states of atoms is 
limited as noted in the next table. No atom is known in which the 0, P, 
and Q orbits are completely filled, and no uncombined atom has more 
than eight electrons in its outer shell. 









\ = K 





2 = L 






0, 1 


2 P 


3 = M 






0, 1 





0, 1,2 




4 = N 






0, 1 





0, 1, 2 




0, 1, 2, 3 



5 = 






0, 1 





0, 1, 2 





0, 1, 2, 3 


5 f 


6 = P 





0, 1 





0, 1, 2 










Wave Mechanics 

Further modifications of the theories of atomic structure followed the 
suggestion by L. de Broglie in 1924 that the electron, like a photon of 
light, had both corpuscular and wave properties, and that wavelengths 
could be assigned to electron motion. This was demonstrated experi- 
mentally in 1927 by C. /. Davisson and L. H. Germer, two American 
physicists, who showed that an electron beam, like X-rays, is diffracted 
by crystals in which layers of atoms act as a diffraction grating (see 
Chapter 13). Since the production of a diffraction pattern can only 
result if the electron beam has wave character, the experiment proved the 
de Broglie idea. The wavelength A of the electron can be calculated from 
the de Broglie equation, 

A = hjmv 

where h is Planck's constant, m the mass of the electron, and v its velocity. 
The new theory developed to use the discovery of the electron's wave 
character and to put the quantum theory on a sounder theoretical basis is 
called wave mechanics or quantum mechanics. The mathematical treat- 
ment of the new mechanics by the physicists E. Schrodinger, W. Heisenberg, 
A. Sommerfeld, and P. A. M. Dirac is extremely complex. 

The electron's motion is pictured in two dimensions and is like that of 
a shaken rope. In the accompanying figure, the wave patterns are shown 
as being limited to those which fit the figure, that is, give reinforced waves. 

FIG. 3-4. Two dimensional representation of electron wave motion : (a) reinforcing 
and (b) non rein forcing waves. Only (a) is permissible. 

A three-dimensional picture is not easy to give and from the student's 
viewpoint the wave concept of electron character suffers because of lack 


of an atomic model, which was one attractive feature of the Rutherford 
and Bohr ideas. Wave mechanics treats the electron's position statisti- 
cally. According to the important Heisenberg uncertainty principle (1927), 
it is impossible experimentally to simultaneously determine the position 
and speed of the electron. It is also impossible to consider the electron 
either exclusively as a wave or a particle, but instead one must think of 
it as having both properties, either of which can become predominant in 
selected experiments. In wave mechanics an electron is said to be in an 
orbital rather than in an orbit. Each orbital can contain a maximum of 
two electrons, each with a different spin. Orbitals have different shapes 
and it will be shown later that this results in predictable geometry of 
molecules and ions whose atoms used certain orbitals in bonding. 

s orhitals are spherical and can be thought of as a shell of charge, the 
density of which is greatest in the center of the shell wall and increasingly 
diffuse away from the center, as the probability of finding the electron in 
different parts of the shell is described. The electron's speed is also 
variable. When both electron motion and speed are treated statistically 
it is found that the Bohr radius and speed are the average values obtained, 

p orbitals arc mutually at right angles to each other and are pictured as 
figure-8 shapes; d and j orhitals are more complicated. (See Fig. 3-11.) 

Electron Configurations 

The combined efforts of physicists in spectroscopic investigations and 
the development of mathematical expressions, and of chemists in relating 
the physical data to chemical properties, ultimately gave us the probable 
electronic structures for all the elements. Since the electrons are used in 
chemical bonding, this information is vital to chemistry. 

The filling of orbitals by electrons, in the progression toward heavier 
elements, can be predicted fairly accurately from two principles. The 
first is that the most stable (lowest energy) orbitals fill in preference to 
others. The second is known as the rule of maximum multiplicity, which 
empirically states that for/?, d, and/subshells which are capable of holding 
6, 10, and 14 electrons, respectively, as many orbitals as possible are 
singly occupied by electrons before electron pairing (by virtue of opposite 
spin) takes place. Orbitals will be represented here by circles, and dots 
inside the circles will represent electrons. Two dots will be a pair of 
electrons with opposed spins and they are all that can be contained in an 
orbital. A single dot will mean an unpaired electron. The following 
figure shows the order in which orbitals are filled. 

Examination of Fig. 3-5 brings forth several generalizations to aid in 
writing ground-state electron configurations. 



8 96 97 98 99 100 lul 

89 90 91 92 93 94 95 
85 86 

81 82 76 77 78 79 80 


56 A 71 72 7l 74 75 64 65 66 67 68 69 70 

- . 

1 3 39 40 41 42 43 


5s CO 

i qT TO i* 26 27 28 29 30 

20 *_.___ : 

vy i 21 22 ?3 24 

2! _______ I 

16 17 18 

L : 

- ^vVjA^ 

13 14 15 
I 1? 

8 9 10 

55 52 53 54 57 58 59 60 61 G2 63 


4 49 50 51 44 45 46 47 48 

FIG. 3-5. Stabilities of electron orbitals. The small numbers are atomic numbers. 
Arrows give the order of filling. 


<1) No electrons enter a </ subshell until the s subshell of the next higher 
shell has been filled. This is the reason that there are never more than 
eight electrons in the outermost shell of any atom. 

(2) The 4f subshell gets no electrons until after the 5^, 5/7, and 6s sub- 
shells have been filled and the 5d subshell has received one electron. The 
same is true of the 5f subshell with respect to prior filling of the 6s, 6/7, 7s, 
and single electron addition to 6d. The consequence of this filling order 
is that there are never more than 1 8 electrons in the next to outermost 
shell of any atom. 

(3) The electron added in each succeeding clement enters the orbital 
with the lowest n + I value, and when two orbitals have that value, addi- 
tion is in the orbital with the lower value of w. 

The student should verify these rules by selecting some examples from 
the figure. 

One may use this diagram to determine the electronic arrangement of 
atoms or ions derived from them, and as will be seen the number and 
pairing of electrons and the type orbitals will be matters for attention in 
predicting such things as the stability, magnetic properties, and geometry 
of ions and molecules. 

Electronic configurations are usually given with a shorthand notation in 
which each subshell is described by main quantum number, electron 
type, and a superscript denoting the number of these electrons. For 
example, element 18, argon, may be described from Fig. 3-5 as having 
2 electrons in the 15- state, 2 in the 2s 9 and 6 in the 2p states; 2 in the 3s 
and 6 in the 3/7 states; or simply, Is 2 , 2s*2p*, 3s*3p*. To describe Mn+ 2 
in the same way, Fig. 3-5 shows that the 0/0/77 has 25 electrons arranged as 
Is 2 , 2s 2 2/> 8 , 35 2 3/? 6 , 4s 2 3d 5 . The dipositive ion has 2 fewer electrons, but 
since the 4s and 3d electrons are approximately equally easy to remove, 
several structures are possible beyond the filled third shell: 4s 2 3d 3 or 
45*34* or 3d 5 . Magnetic measurements (described later) indicate five 
unpaired electrons, thus only the last two are possible configurations. 

The Periodic Table and Atomic Structure 

The periodic arrangement of elements may profitably be reviewed at 
this point in the light of electron configurations. Figure 3-5 and Fig. 3-6 
should be referred to in this discussion. 

The first orbital, Is, can only contain two electrons and the K shell is 
completed in helium. The next orbital to begin filling is the 2s in lithium. 
The next electron completes the 2s orbital in beryllium. The next three 
electrons to be added in B, C, and N appear singly in the 2p orbitals, and 
pairing of these occurs in the next three elements, O, F, and Ne. As 
previously explained the L shell is complete with two s and six/7 electrons, 
















therefore the next electron to be added in sodium, element 11, is a 3^ 
electron. The 3s and 3p orbitals are built up in the same fashion as the 
2s and 2/7, but ten more electrons are needed to complete the 3d orbitals 
and this is not accomplished until Cu. From Fig. 3-5 it is seen that the 
4s level is of slightly lower energy than the 3d so it is filling in K and Ca. 
As spectroscopic study shows, the next electrons go into the five 3d 
orbitals, between Sc and Cu, rather than the 4p positions. The elements 
in which this takes place constitute the first group of transition elements 
(see Chapter 17). None have all electrons paired. The 4s electrons 
remain one or two in number while the inner 3d orbitals are filling. In 
Cr and Cu a 4s- electron has dropped back to the 3d state, this being 
possible since those two levels have about the same energy. 

The 3d orbitals are filled and the M shell is closed with 18 electrons at 
element 30, zinc. From Ga to Kr the three 4p orbitals are filling. The 
next two electrons go into the 5,y orbital, as shown in Fig. 3-5. These are 
elements 37 and 38, rubidium and strontium. Beginning with number 39, 
yttrium, and extending through number 47, silver, 4d orbitals rather than 
5p arc filling and these elements comprise the second transition series. 
The elements between cadmium and xenon have the 5p orbitals filling 
rather than 4J\ and this is completed with xenon. Cesium and barium 
have the xenon core of electrons plus one and two 6s electrons, respectively. 
The next clement is lanthanum and its added electron is in a 5d orbital. 
The next fourteen elements are peculiar in this way: electrons are added 
in the 4f orbitals so 4f, 5</, and 6p are all incomplete. These elements 
are called the lanthanide series of rare earths and are grouped separately 
at the bottom of the main table in Fig. 3-6. By the time element 71, 
lutecium, is reached, the N shell is fully expanded from 18 to 32 

Following the lanthanide series is another transition group, Hf through 
Au, in which the 6s level holds one or two electrons and filling is in the 
five 5d orbitals. From 81, thallium, to 86, radon, the three 6p orbitals 
are being completed. 

The next horizontal row in Fig. 3-6 starts as the preceding one, one 
electron in an s orbital (Is) for francium and two for radium. The next 
added electron in actinium goes into a d orbital (6d) just as happened with 
clement 57 in the previous row. Beginning with 90, thorium, there are 
three incomplete orbitals, 5f, 6d, and Is. Of these, 5f is filling while 6d 
contains one, and Is two electrons. The elements 90-103, of which the 
last two are as yet unknown, constitute a second rare earth group known 
as the actinide series. The elements beyond uranium are not naturally 
occurring but have been prepared synthetically by nuclear transformations. 
They are known as the transuranium elements. 



Atomic Size and Chemical Properties of Elements 

One might suppose that atomic size will vary directly with atomic 
number since the number of electrons increases and electron shells are 
filling. This is true for a given vertical group of elements in the periodic 
system but not true when considering a horizontal period. In the former, 
a new shell is being filled which results in a somewhat larger radius, but 
in the latter case the quantum group remains constant as the atomic 
number (nuclear charge) is increasing. This means that electrons are 
drawn closer to the nucleus and the size, generally speaking, shrinks. It 
increases suddenly with the alkali metals of group I A, however, because 
a new shell is starting (Fig. 3-7). 

ol I I I I I 

40 50 60 

Atomic number 

FIG. 3-7. Covalent radii of the elements. 

Diffraction studies and quantum mechanical calculations have given 
approximate radii of elements in compounds in which bonding is by 
electron sharing (covalence) and radii of ions in compounds in which 
bonding is electrostatic between the ions (electrovalence). A number of 
generalizations can be made from a study of these radii, as noted in the 
next paragraphs. 

A postive ion like Na 4 " is expected to be smaller than the corresponding 
atom, because only the K and L shells are left after the electron is removed 



from the M shell. On the other hand the radius of a negative ion like 
Cl" will be slightly larger than the corresponding atom, since the added 
electron is not accompanied by an added proton in the nucleus. Chloride 
ion has the same electronic configuration as argon but with one less unit 
of nuclear charge is more expanded. For a given element, a cation formed 
from the neutral atom is always smaller than the atom, whereas the 
corresponding anion is always larger. If several oxidation states are 
studied, the ionic size is found to vary inversely with the positive oxidation 
number, as shown in Table 3-4. 


Group IA 

Group VA 

No. 3 


0.60 A 

No. 7 

N+ 5 

0.11 A 




N~ 3 






As+ 5 





As+ 3 






Sb+ 5 


Sb- 3 


Group HA 


Be' 2 


Group VIA 


Mg+ 2 
Ca+ 2 
Sr+ 2 
Ba+ 2 




S+ 6 

s- 2 


First Row 

Group VIIA 






Cr+ 3 



C1+ 7 


* Cr 6 





Mn+ 2 



Br^ 7 


Mn+ 7 





Fe+ 2 



I+ 7 


Fe+ 3 




Figure 3-7 shows that changes of atomic size with increasing atomic 
number are not great in the transition elements. The reason for this is 
that electrons are not filling the outermost 4p orbitals, but 3d instead. 
The two rare earth series at the bottom of the table show decreases in 
sizes with increasing nuclear charge, since in both series the filling is two 
orbits in from the outside, and each increase in atomic number merely 
results in a tighter holding of all electrons. These size changes are known 
as the lanthanide and actinide contractions. 

Because of the regular decrease in diameter of the +3 lanthanide ions, 


they have been used in several classical studies to test trends in properties 
as functions of ionic size. As an example, it has been shown that stability 
of higher oxidation states increases in the series with increasing atomic 
number. It is characteristic of the higher states to have electron shared 
bonds (covalence) rather than ionic bonds, since highly charged ions are 
not stable. As the size of these ions varies inversely with atomic number, 
the heavier elements should hold electrons more tightly, show more 
covalence, and have more compounds containing the metal at higher 
oxidation numbers. Such trends have been experimentally verified. As 
another example, one expects the hydroxides, M(OH) 3 , to become less 
basic with decreasing ionic size, since the smaller ions will exert greater 
electron attraction and allow H + but not OH~ to dissociate. This 
increase in acidic character has been observed. La(OH) 3 for instance is 
not only the strongest base in the series but also the strongest trivalent 
base known. 

More quantitatively, if the ionic radius of the metallic element is 
< 0.5A, its hydroxide is acidic; if slightly > 0.5, the hydroxide is weakly 
acidic in the higher oxidation states; if 0.5 0.9, the hydroxide is 
amphoteric; and if the radius is > 0.9, the hydroxide is definitely basic. 

Ionic sizes and charges determine the orientation and number of 
groups (coordination number) that are capable of being held about a 
central element. The first horizontal row of elements has a coordination 
number of 3 with respect to oxygen, as shown by ions like CO^ 2 and NO;j". 
These are planar. Second- and third-row elements, being larger, can 
coordinate four oxygens giving tetrahedral ions like POj~ 3 and ClOj". 
Heavier ions can coordinate six oxygens in an octahedral arrangement, 
as in Sb(OH)^*. Coordination numbers are considered further in the 
next chapter. 

lonization Potentials 

Within a (vertical) family of elements the outer electron levels in the 
larger atoms are less affected by nuclear attraction than outer electrons in 
the smaller atoms, since ihe former are further removed from the nucleus. 
This effect is manifested in the ionization potentials of the elements (see 
Fig. 3-3 for H). A plot of atomic number versus ionization potential 
shows maxima corresponding to the noble gases and minima for the alkali 
metals, as expected. In addition, the potential varies inversely with 
atomic size within a family of elements, which is also expected for reasons 
given before: smaller atoms hold electrons more tightly, since the center 
of positive charge is closer to them. 

In the series Li-Cs, cesium has the lowest ionization potential (removal 
of a 6s electron easier than a 2s in the lithium atom), is therefore the best 


reducing agent, forms a positive ion the easiest and has the greatest 
metallic character (Chapter 13). 

In the series F-I, the electron is expected to be most difficult to remove 
from the smallest atom, since it must be taken from a shell closer to 
nuclear attraction than is true for the heavier atoms, and the potentials 
bear this out. The order of ionization potentials is the same as the trend 
in electron attracting abilities of the halogens: fluorine is the most 
electronegative element known and hence an excellent oxidizing agent. 

Removal of a second electron from a unipositive ion requires more 
energy than removal of the atom's first electron, since it must be done 
against the attraction of a positive ion. Further electron removals become 
progressively more difficult, precluding the formation of highly charged 
positive ions in simple chemical reactions. If one electron is readily lost, 
but not two (Li for example), the valence of the element is + 1, if two are 
detachable with reasonable energy, but not three (Ca for instance), the 
ion's valence is +2, etc. Table 3-5 helps such speculation. 

By examining the ionization potential variations in proceeding hori- 
zontally across the periodic table one notices a general increase in value, 
since the additional electrons in heavier atoms are at about the same 
distance from the nucleus, thus the increased nuclear charge is capable of 
holding all electrons more tightly. There are some exceptions. The 
apparent discrepancy in the aluminum value (5.98) compared to that of 
the preceding element, Mg (7.64) (Table 3-5) is due to the fact that the 
added electron in Al is in a 3/7 orbital which is shielded by the just com- 
pleted 3s orbital containing two paired electrons. This shielding effect 
means that the electrons closer to the nucleus absorb a disproportionate 
share of nuclear attraction, repel outer electrons, and in general allow the 
latter to be more capable of escape. Tn a given quantum shell, electron 
energies increase in the order s < p < c I < /. It is thus easier to remove 
an /electron than an s, p, or d electron of the same shell. 

In summary, one may state that ionization potential varies with nuclear 
charge, atomic radius, electron orbital, and screening effect. Generally 
the ease of electron removal decreases with increasing atomic number in a 
given horizontal period and increases with increasing atomic number in a 
given vertical column of the periodic table. 

Trends in Families of Elements 

Chemical reactions involve electrons in the outermost shells, and so 
their number, grouping, and energy are important in determining chemical 
properties. Study of the preceding sections and accompanying tables and 
figures shows that a repetition of outer electron configuration occurs 
periodically, and elements having such similarity possess chemical 






lonization Potentials In Electron Volts 































































































































































































































properties in common. This idea is stated as the periodic law, which says 
that the chemical character of elements is a periodic function of their 
atomic numbers. The purpose of the following sections is to point out 
some trends both in horizontal (period) and vertical (family or group) 
progression through the periodic chart. 


This group of six "noble" gas elements has almost no chemical properties, 
and it was this feature that interested chemists. Once the electronic con- 
figurations were deduced, theories of chemical bonding followed, using the 
idea that other elements tend to lose, gain, or share electrons to attain the 
nearest noble gas (closed) electron arrangement, since that is the one with 
greatest stability and lowest energy. Table 3-6 shows that except for 
helium each element has eight electrons in the last shell. 


Shells and Electrons 


7< If*mpnt > 









































Groups IA and IB 

Hydrogen is often included in both IA and V1IA, because like IA 
elements it can lose an s electron and become a positive ion or like VIIA 
elements can gain an electron and become a negative ion (H~~ is the 
hydride ion). The other members of IA are known as the alkali metals. 
All have low first-ionization potentials, hence all give univalent positive 
ions and act as strong reducing agents. The ions have noble gas electron 
configurations and form ionic compounds bonded by electrovalences (see 
p. 51). As in all periodic groups, the first element of the series is the one 
which differs most from the rest, whereas the heavier elements are generally 
the most closely related. For example, lithium has the highest melting 
point, boiling point, and hardness and shows some properties characteristic 
of IIA elements. Other chemistry of these ions is discussed in relation 
to their qualitative analysis in Chapters 19 and 20. 



The elements in group IB, Cu, Ag, Au, are the subgroup, whereas the 
alkali metals are the main group of group I elements. Main group 
elements are characteristically close in electron configuration to noble gas 
atoms, whereas subgroup elements are not, making for a single valence state 
in the former and variable valence in the latter. Elements of both groups 
IA and IB have one electron more readily removable than the rest, 
although the electron is more difficult to remove from IB members, and 
they accordingly show some electron sharing (covalent) type bonding 
which is generally lacking in compounds containing IA elements. The 
penultimate (next to last) electron shell in IB elements contains 18 electrons, 
and some of these can be used in bonding, giving the variable valence 
characteristics. The penultimate shell of 1 A metals contains two electrons 
in Li and eight for the rest; these groupings are too stable to be chemically 
reactive. Further chemistry of copper is given in Chapter 16 and that of 
silver in Chapter 15. 


Shells and Electrons 




















































Groups IIA and IIB 

The elements of IIA, Be, Mg, Ca, Sr, Ba, and Ra, are known as the 
alkaline earth metals. Each has two s electrons in the outermost orbital 
and each exhibits divalency. Due to its small size, Be is the most 
"different" element of the series. Like other small ions, Be+ 2 is best 
capable of forming complexes and exhibiting covalence. Be resembles 
Al of group IIIA in this respect, something other group IIA elements do 
not. The gradation of properties due to change in ionic size are those 
expected; lower oxidation potential, etc., with heavier members of the 
group. Though the penultimate shell contains 18 electrons, none are 
available for bonding as described for IB metals, since IIA ions are 


dipositive, and too much energy is required to remove a third electron 
(Table 3-5). 

The group IIB elements, Zn, Cd, and Hg, are- also divalent since their 
electron structures are similar to those of group II A metals. Mercury 
is unique among the IIB group. It forms the mercurous ion ^Hg:Hg+, 
in which one electron from each atom is lost and another is used in a 
covalent bond. Covalency is more common in the B group than A group. 

Groups IIIA and 1IIB 

The III A elements are B, Al, Ga, In, and Tl. The last shell of electrons 
includes a completed s orbital and one electron in a p orbital. The 
oxidation states of these elements are (I), (II), and (III), although much 
of their bonding is covalent. Boron forms no simple ions like B i3 and 
is always covalent. Boron resembles elements adjacent to it, Be and C, 
almost as much as it does elements below it. The heavier members of the 
group arc more metallic. A glance at the properties of their hydroxides is 
instructive. B(OH) :l is weakly acidic and is usually written as boric acid, 
H 3 BO 3 (see Chapter 22, part I). A1(OH) ; , dissolves to give salts in either 
acid or base and hence is amphotcric; aluminum chemistry is discussed in 
Chapter 17. T1(OH) 3 has no acidic properties and Tl(OH), the only 
stable univalent group III hydroxide, is a fairly strong base. As a rule, 
base strength of hydroxides varies inversely with the oxidation state of the 
metal and varies directly with increasing atomic weight in a family oj 
elements. All the metals have co-ordination numbers of 4 or 6 in group III. 

The group IIIB elements include Sc, Y, La, and Ac, and the two rare 
earth series as well (see previous discussion). The electronic structure of 
the latter has been given and it will suffice here to say that the normal 
oxidation state is (III), and changes in chemical properties arc generally in 
the order expected through the series. Scandium resembles aluminum 
in that its salts are hydrolyzed and Sc(OH) :j is very weakly basic. Lan- 
thanum salts arc less hydrolyzed, its hydroxide is fairly basic, and its 
chemistry is something like that of Ca. The bonding in these elements is 
predominantly covalent and they have coordination numbers of 6 
and/or 8. 

Groups IVA and IVB 

All members of these groups have 4 electrons available for bonding and 
by means of electron sharing can acquire stable octets. The elements 
of IVA are C, Si, Ge, Sn, and Pb. Again the difference in properties of 
the first two elements is great, and that of the lower elements much less. 
Carbon can hold a minimum of 2 and maximum of 4 atoms in covalence, 
whereas the other elements can bond with 2, 4 or 6 groups. Carbon is 


unique among all elements in that C C bonds are strong and long chains 
and extended rings containing many C atoms are known (Chapter 12). 
Silicon and germanium also have the properties of nonmetals; they do 
not give simple ions in solution but are capable of forming covalently 
bonded compounds, many of great complexity. Lead and tin are 
metallic; they are discussed in Chapter 16. 

The TVB elements, Ti, Zr, and Hf, have 2 electrons in the last level and 
10 in the penultimate shell and have oxidation states of (II), (III), and 
(IV). As expected, covalency decreases with increasing atomic number. 
None of these is capable of existing as an ion with +4 charge in aqueous 
solution, however, since the distorting effect of an ion with such a high 
charge density means it will react with water to form oxygenated ions such 
as TiO+ 2 . (See Chapter 20.) 

Groups VA and VB 

The members of group VA are N, P, As, Sb, and Bi and of VB, V, Nb, 
and Ta. There is not much similarity between the two groups except for 
some compounds whose oxidation states are (V), as POj 3 and VOj :j . 

The main group elements form a well-defined scries with regularly 
changing properties due to atomic size variation. The smaller atoms, 
having greater influence over electrons, can form stronger acids like 
HNO ; , than can larger central atoms, whose weaker hold on electrons per- 
mits the electrons to in turn bond acidic hydrogens more iirmly. Multiple 
bonds and general covalency are more common again in the lighter 
elements. Sb and Bi are the only metals in group VA and they can 
form trivalent cations; pentavalent ions have too great a charge to exist 
and this oxidation state is only known in complex ions like Sb(OH);r. 
See Chapter 16 for further chemistry of As, Sb, and Bi. 

The group VB elements are among the transition metals and show the 
expected gradation of properties. Niobium and tantalum are chemically 
quite similar, whereas vanadium is the "different" group VB metal. Us 
chemistry is complex (see Chapter 20) because of the importance of all 
its oxidation states, (II), (III), (IV), and (V). These, respectively, have 
3, 2, 1, and unpaired electrons. 

Groups VIA and V1B 

The A group elements are O, S, Sc, Te, and Po and the B group are 
Cr, Mo, and W. Except for some compounds in which the oxidation 
state is (VI), SOj 2 , TeO' 2 , CrO^ 2 , WO 4 2 , there are few similarities between 
the groups. // is a rule that the dissimilar it v increases between main and 
subgroup as elements further toward the right of the periodic chart are 



The group VIA elements need 2 electrons to complete an octet in the 
last shell and do exist as ions having a valence of 2; the members 
beyond oxygen also show oxidation numbers of (III), (IV), and (VI). 
Bonding is covalent in these and none of the elements is markedly metallic. 
Oxygen has the greatest electronegativity among them (Fig. 3-14), and is 
different from the others because of this. 

Of the group VI B elements, Cr is unique (Chapter 17) and Mo and W 
more similar. Cr TIT , Cr VT , Mo VI , and W V1 are the important oxidation 
states and many complex ions containing them are known. The tendency 
toward condensation of oxygenated compounds and ions increases with 
atomic number (simplest: 200^ + 2H+ = Cr 2 O 7 2 + H 2 O), which 
makes the chemistry of Mo, and more especially W, exceedingly complex. 
Acid strength decreases with increasing condensation^ a generalization not 
restricted, however, to these elements. Mo and W are considered further 
in Chapter 20. 

Groups VII A and VIIB 

In group VIIA are the closely related halogens, F, Cl, Br, and I (and 
At), and in VIIB, Mn, Tc, and Re, whose similarity to the main group is 

The halogens are electronegative, since by gaining an electron each 
attains the next rare gas configuration. Besides a valence of 1, these 
elements show an increasing number of covalences with increasing atomic 
number; F having a maximum of 2 as (HF) 2 ; Cl, 4 as HCIO 4 ; Br, 5 as 
BrF 5 ; and I, 7 as IF 7 . F is the "different" halogen. It is a very powerful 
oxidant, has the highest electronegativity of any element, and because of 
this and its small size forms compounds the other halogens cannot. 


Shells and Electrons 








































Of the group VIIB elements, only Mn is important; Re is rare, and 
Tc is one of the elements only known through synthesis by nuclear 


transformation, in this case deuteron bombardment of Mo followed by 
chemical separation. It resembles Re more closely than Mn. Manganese 
has all positive oxidation states of (1) through (VII), but the only ion it 
forms that is related to those of the halogens is MnOj, which like ClO^ 
has the central element in the (VI I) state. Manganese chemistry is more 
fully described in Chapter 17. 

Group VIII 

This central periodic grouping contains three sets of three elements 
each. For purposes here the first member of each group will be discussed 
briefly, namely Fe, Co, and Ni (note their positions, Fig. 3-6), since they 
are included in the laboratory work. (See Chapter 17.) 

With increasing atomic number, the ability to form the tripositive state 
decreases sharply; Fe 111 is common, quite a number of Co 111 complexes 
have been prepared, but only a few compounds of Ni IlT are known. Salts 
of the divalent ions are most common and stability of this oxidation state 
increases in the same order. Oxidation states beyond (III) are not stable 
in neutral aqueous solution because of their strong oxidizing power. These 
typical transition metals have unpaired electrons due to an incomplete 
penultimate shell containing between 8 and 18 electrons. Solutions of 
such ions absorb light in the visible region and appear colored. Iron has 
8 more electrons than the preceding rare gas, cobalt 9 more, and Ni, 10. 
The heavier elements below these three show similar trends. In the 
triad Ru, Rh, Pd, the tendency for formation of higher oxidation states 
decreases in that order as is the case for the triad Os, Ir, Pt. Thus in the 
fluorides of the latter, OsF 8 is known but only IrF 6 and PtF 3 have been 

Chemical Bonding 

Now that ideas concerning electronic configurations have been reviewed 
and extended and some relationships among periodic groupings of 
elements mentioned, the remainder of this chapter will deal with forces 
holding atoms together and methods of investigating chemical bonding. 


The electrovalent or ionic bond concept was given by Kosscl in 1916. 
He simply suggested that properties of electrolytes could be best explained 
if it is assumed that salts are composed of ions bonded together by electro- 
static attraction between unlike charges. Ions are formed by electron 
transfer (redox) between atoms such that the resulting configurations have 
greater stability. Anions formed in this way have rare gas structures, 


whereas cations may or may not result in such electron arrangement. The 
simplest cases are those involving elements of the electronegative groups 
VIA and VIIA and the electropositive groups IA and IIA. Figure 3-8 
shows how typical elements of these groups may react to yield noble gas 

2Li- + b". = 2Li + 

Mg: + 2"Ci: = Mg^ + 2:C1-" 

FIG. 3-8. Formation of two compounds by electron transfer. Dots represent 

electrons from the metals, x 's those from the nonmetals. The ions are held together 

by electrostatic (coulombic) forces exerted in three dimensions which result in the 

building of symmetrical ionic crystals. 

Subgroup elements form ions that do not have noble gas arrangements. 
Subgroup ions are characterized (a) as possessing 18 electrons in the last 
shell or (b) as having between 8 and 18 electrons in the last shell, which is 
called the transition type structure. Thus Zn, 2, 8, 18, 2, gives Zn+ 2 , 
2, 8, 18, and Cu, 2, 8, "l8, 1, gives Cu 1 , 2, 8, 18, while Fe, 2, 8, 14, 2, gives 
Fe i2 , 2, 8, 14, and Fe+ 3 , 2, 8, 13. The most stable cations are those with 
the rare gas structure, next are those with the 18 electron group arrangement, 
and last are those of the transition ion type. The latter types increase 
their stabilities, however, through complex formation by addition of 
groups capable of donating otherwise unshared electron pairs to vacant 
orbitals in the metal ions (Chapter 4). 

From a study on bond types as related to atomic size, ionic charge, and 
electron configurations, K. Fajans (1924) announced several generalizations 
that aid in predicting what kind of bonding to expect in a given compound. 
According to the Fajans Rules, electrovalency is expected if (a) the ions are 
electronically stable (as explained above), (b) the ionic charges are small, 
and (c) the cation is large and the anion is small. The reason for the 
operation of these principles is that electron loss is best accomplished if 
only one electron is removed and its position was originally distant from 
the nucleus (compare Li and Cs, Table 3-5, for example), and electron 
gain is best made by small atoms in which the attracting nucleus is close. 
If the compound in question has characteristics opposite the above, 
electron shared bonds (covalence) is expected. There are many inter- 
mediate cases as well in which bonds are said to possess a mixture of 
electrovalent and covalent character. 


Ionic compounds are strongly bonded. They can be thought of as 
large symmetrical aggregates of ions in which a given ion is attracting 
and is surrounded by a fixed number of oppositely charged ions. As a 
consequence the compounds show electrical conduction in the molten 
condition as well as in aqueous solutions, and the crystals are hard and 
high melting. 

With enough data available one can calculate semiquantitatively whether 
or not ionic bonds are capable of formation when two free atoms combine. 
Though this will not be developed further here, one may say first that if 
the energy of the ionic compound is less than the energy of the starting 
atoms, bond formation is possible. One then takes into account energy 
terms related to the heat of reaction and to the production of free atoms 
from the solid or gaseous elements. Other factors to consider include 
the cation's ionization potential, the tendency for the electronegative 
element to gain electrons ("electron affinity"), and the attractive and 
repulsive forces acting between the resulting ions when the crystal forms 
(crystal energy). Calculated and measured values for various of these 
terms show good agreement. 

Magnetic Susceptibility 

Ionic compounds have been studied in many ways, some of which are 
given in Chapter 5 (ions in solution) and Chapter 13 (symmetry in crystals). 
An interesting field of study that will be outlined here for its application 
later in the discussion of Werner ions is that of magnetic susceptibility. 

Matter is ferromagnetic (limited to materials ordinarily considered 
magnetic, like Fe, Co, Ni, and their alloys), paramagnetic, or diamagnetic. 
The latter effects are much weaker than ferromagnetism. A paramagnetic 
substance in the solid state or in solution will move in the direction of 
increasing field strength of an electromagnet, whereas a diamagnetic 
substance moves out of the magnetic field. Measurements of these 
effects are made by use of the Gouy balance (1889). 

It is a fact that substances containing atoms having unpaired electrons 
are paramagnetic and others are diamagnetic, and, further, that their 
magnetic moments are directly related to the number of unpaired electrons. 
The rotational motion of unpaired electrons sets up a weak but measurable 
moment or field while paired electrons cancel each other's effect. The 
balance pictured in Fig. 3-9 is capable of distinguishing these effects and 
their magnitude. From Gouy data the moments can be calculated and 
are expressed in Bohr magnetons, the unit being 9.18 x 10~ 21 gauss-cm 
per ion. An ion having x unpaired electrons will have a calculated moment 
of [x(x + 2)]* magnetons. From Fig. 3-5 and magnetic data, one can, for 
example, deduce the rule of maximum multiplicity. It will be shown in 




FIG. 3-9. The Gouy balance for determining magnetic susceptibilities. If the sample 
is paramagnetic it is drawn toward the magnet and more counter weight is added on 
the balance. Jf it is diamagnetic, the field repels the sample and less counter weight 

is needed. 














K 1 



















Fe m 










the next chapter how magnetic data permit the prediction of complex 
ion geometry since the deduced electron combinations are known to have 
certain symmetrical spatial distributions. 


The postulate that atoms may react to form more stable electron distri- 
bution through sharing rather than actual transfer of electrons was made 
by (7. N. Lewis in 1916. He defined a covalent bond as a pair of shared 
electrons counting as part of the outer shells of two atoms. This is a 
simple idea but a very important one. Instead of ions forming and being 
held together by electrostatic attraction, covalent compounds are charac- 
terized by the formation of molecules whose atoms arc bonded together by 
strong, shared electron-pair bonds. Molecules are bonded to other 
molecules by weaker bonds called ran <ter Waals forces, after the man who 
first discussed them quantitatively. These cohesive forces are a manifesta- 
tion of attraction of atomic nuclei for electrons of neighboring atoms. 
Attraction between molecules is directly related to the number of electrons 
available per molecule but is considerably weaker than other bonding 
discussed th is f <r, since there is a substantial compensating force of repul- 
sion on a eiven atom's electrons by electrons of neighboring atoms. 
Crystals formed in this way (solid group elements, halogens, most 
organic compounds, etc.) are called molecular crystals to differentiate 
them from most salts that form ionic crystals. As a consequence, typical 
covalent compounds arc insulators and generally give soft crystals having 
low melting and boiling points. 



Elect. Cond. 
(liquid state) 


M.p., J C 

B.p., C 






MgF 2 





A1 2 3 





NH :i 








AsF 5 




Simple representations showing only the outermost electrons are again 
convenient in visualizing covalent bond formation. The symbolism 


usually used is to designate each covalent bond as a line and electrons 
not used in bonding as X's and dots. Electron shared bonds are known 
containing 1, 2, 3, 4, and 6 electrons. The 2-, 4-, and 6-electron bonds are 
the important ones and are respectively called single, double, and triple 
bonds. The strength of the bond increases and the distance between 
atoms decreases in the order given. 

(a) :CUCi: or :C1 C1-* 

(b) :O::C.":b: or :6=C=O: 

(c) :N:;:N: or :N==N: 

FIG. 3-10. Three types of covalent links: (a) single (b) double (c) triple. Note that 

each atom attains a noble gas electron arrangement via equal electron donation and 

sharing, and that some outer orbit electrons are not used in bonding. 

With larger atoms, the requirement that a noble gas configuration be 
formed (the octet rule) cannot be maintained since compounds like 
IF 6 , SF fi , IF 7 , and OsF 8 are known. Since two 2s and three 2p orbitals 
are available in elements Li through F, their maximum covalency is 4; 
in elements Na through Br, one 3s, three 3/>, and five 3d orbitals, a total of 
9, are theoretically available, but maximum covalency is 6; in still heavier 
atoms the maximum observed covalency is 8. 

Bond Orbitals and Covalence 

One important restriction on the formation of covalent bonds which 
has not been explained as yet is that only those atoms combine which 
have at least one orbital containing a single or unpaired electron. The 
bond then results in a pairing of electron spins. For instance, the 
electronic structure of nitrogen can be written as b 2 , 2s 2 2p2p2p, showing 
three unpaired 2p electrons. Three electrons of opposite spin can enter 
these orbitals to form three equivalent covalent bonds. In the formation 
of ammonia, pictured below, these electrons are contributed by three 
hydrogen atoms. 

It was mentioned earlier that while s orbitals are spherical or non- 
directional, p orbitals are figure-8 shaped and at right angles to each other. 
Since the axes of three-dimensional space are usually designated #, y, and 
z, p orbitals are descriptively labeled in like fashion. For example, the 



electrons of the N atom are often designated Is 2 , 2s 2 2p x 2p y 2p z . In Fig. 
3-1 1 there is pictured an overlapping of s and/? orbitals to effect covalent 
bonding by pairs of spin compensated electrons. 


FIG. 3-11. (a) One of the three 2p orbitals of a nitrogen atom, occupied by a single 
electron, (b) The \s orbital of a hydrogen atom occupied by a single electron, (c) 
One of the three sp orbitals of NH 3 formed by overlapping of an s and p orbital and 
containing two electrons. The dots are the atomic nucleii. 

Carbon is a special case. The structure of the atom is \s 2 2s 2 2p2p, 
indicating it should be able to form only two covalent bonds, as in carbon 
monoxide, C=O. A 2s electron is capable of promotion, however, to 
the higher energy 2p state, giving Is 2 2s2p j .2p y 2p z , the energy for this coming 
from that released when the extra two bonds form. Investigation of 
compounds like CH 4 , CC1 4 , etc., shows that all 4 bonds are equal in 
length and strength despite an expected difference in one bond, since it 
presumably involves use of a 2s electron while the others are in the 2p 
condition. Pauling, Slater, Mul liken, and others have shown with 
quantum mechanics that what happens is an amalgamation (hybridization) 
of the 4 orbitals to give 4 equivalent bonds orientated in space to the 
corners of a regular tetrahedron. This is designated as sp 3 bonding, 
since one s and three p electrons are involved, and is always tetrahedral. 
All the steriochemistry of carbon bears out this as do measurements of 
carbon bond angles by electron diffraction; 109 28' is the tetrahedral 
angle. Hybridization of other orbitals, particularly in the transition 
metals, is explained in the next chapter. The sp 3 bonds are unsymmetri- 
cally figure-8 shaped, directed toward the corners of a tetrahedron 


Another feature of electron-shared bonds is the concept of resonance, 
introduced by Heisenberg in 1926 and widely applied since then by 


Pauling and others. For our purposes, it will be sufficient to say that the 
properties of some substances arc best accounted for by the writing of 
several possible electronic structures whose combination gives a more 
satisfactory representation than does a single structure. One can think 
of electron pairs moving very rapidly among the positions pictured in the 
usual electronic structures. The actual condition is a single resonance 
hybrid or composite, and it is a consequence of wave mechanics that this 
system has lower energy than otherwise expected and is said to be addi- 
tionally stabilized to the extent of the resonance energy of the system. To 
aid in understanding resonance there are many analogies that can be 
made with common experience, though they arc all oversimplified. For 
example, a train (electrons) making continuous round trips among several 
stations (conventionally pictured electron positions) spends most of its 
time somewhere among the stations, and if one were asked to draw a 
picture representing an average condition on this railroad he might draw 
several diagrams showing the train in each of several locations and then 
say that the average condition is a combination of all the pictures. If 
the electrons are considered moving too rapidly to be thought of as 
individual particles but rather as a zone of charge, then this further analogy 
is useful: an orange paint (the hybrid) can be made by mixing yellow and 
red paints (individual electronic states). 

As one example of a resonating molecule, ozone, O ; ,, contains two 
oxygen-oxygen bonds which by electron diffraction measurements are 
shown to have equal lengths. If one draws the electronic structure for 
ozone completing an octet for each oxygen, however, the result is one 
double and one single bond, which does not explain the experimentally 
found bond length. By postulating resonance between two such hypo- 
thetical structures one says that their composite or hybrid structure is 
capable of representing the facts. 

9 ' < 

FIG. 3-12. Two hybrid structures of ozone. Curved arrows represent resonance by 

electronic shifts in which electron pairs are moved to picture possible states which 

contribute to the composite. A comma or a double-headed arrow is used between 

the hybrids to indicate their interconvertability. 

Chemists have used these ideas widely, particularly in the formulation 
of organic reaction mechanisms that are explained on the basis of 
resonating molecules and ions possessing special energy and structural 


Partial Tonic Nature of Covalency 

Between the extremes of electrovalency like Cs + F~ and covalency like 
H : H is considerable room for transition, and most chemical bonds are 
now known to have mixed characteristics. The resonance theory explains 
this by showing the possibility of contribution by both covalent and ionic 
structures to the hybrid structure. Unless the compound is a binary 
one of a group IA or IB metal and a halogen, oxygen, or sulfur, it is 
conventional to represent it as a covalent molecule with the understanding 
that the covalent bonds have partial tonic character. 

(a) (b) 

FIG. 3 13. Two structures which help explain the properties of hydrogen bromide: 

(a) covalent and (A) ionic. The H Br bond is calculated to be about 90? t ; covalent. 

In water, the bond is broken and HBr becomes a strong electrolyte. 

The Electronegativity Table 

An aid in estimating the ionic contribution in otherwise covalent bonds 
has been afforded by Pauling in his development of a table of electro- 
negativities. These numbers, derived from measured heats of formation 
but put on an arbitrary scale from 0.7 to 4.0, give the relative tendencies 
of atoms to attract electrons in a covalent bond. The following generaliza- 
tions and uses can be made of the compilation in Fig. 3-14. 

(1) The larger the electronegativity value, .r, the more powerfully that 
atom attracts electrons and tends to become a negative ion. 

(2) The larger the difference between the electronegativities of atoms 
A and B, the greater is the ionic character of the bond. Pauling has 
calculated that if the difference, .r A j- H , is zero units, the ionic character 
is zero per cent; 0.2 units corresponds to 1%, 0.6^7%, 1.0?- 18%, 
1.4^32%, 1.8 c* 47%, 1.9-^50%, 2.2^,61%, 2.6^74%. 'if the 
bond is essentially covalent or ionic according to this, it is written accor- 
dingly. Thus for sodium chloride, x x .r, 4 = 2.1, indicating that the 
compound is predominantly electrovalently bonded. 

(3) The larger the electronegativity difference, X A .r,,, the greater 
generally is the heat of formation of AB and the more stable is the 

(4) An electronegativity of less than about 1.8 is a characteristic of 
metals; amphoteric elements and hydrogen have a value of about 2 and 
nonmetals have values above 2. 



Li Be B C N O F 
1.0 1.5 2.0 2.5 3.0 3.5 4. 

Na Mg Al Si 
0.9 1.2 1.5 1.8 

P S Cl 
2.1 2.5 3.0 

K Ca Sc Ti As Se Br 
0.8 1.0 1.3 1.6 2.0 2.4 2.8 

Rb Sr Y Sn Sb 
0.8 1.0 1.3 1.7 1.8 

Te I 

2.1 2.4 

Cs Ba 

0.7 0.9 





Electronegativity scale 

FIG. 3-14. Pauling's electronegativity table. (Adapted by permission from Lmus 
Pauling, Nature of the Chemical Bond, Cornell University Press, Tthica, New York, 


Coordinate Covalence 

A coordinate covalent bond is formed when one atom or ion (donor), 
with at least two valence electrons not being shared, donates an electron 
pair to another atom or ion (acceptor) capable of holding them. This 
capability stems usually from an incomplete octet of valence electrons in 
the acceptor. Once formed, this bond is equivalent to a covalent bond. 
Examples and further consequences of coordinate covalence are given in 
Chapter 4 (coordination compounds) and Chapter 5 (Lewis acid-base 

Dipole Moments 

Experimental evidence of electronic displacement in bonds and asym- 
metric electron distribution in molecules is gained by the determination of 
dipole moments, as pointed out in 1912 by P. Debye. The magnitude of 
the dipole moment // of a bond A B is dependent upon the electro- 
negativities of A and B and is the product of either of the electronic 
charges e and the distance / between them, ju = el. Dipole moments are 
approximately additive (vector quantities acting in the direction of the 
bond) so that an unsymmetrical molecule will have a permanent dipole 
(be a polar molecule), whereas a symmetrical one will not have a dipole 
moment and will be a nonpolar molecule. Molecules of polar compounds 



attract each other and hence show mutual solubility, and the pure sub- 
stances have lower volatility than otherwise expected, since energy is 
needed to overcome intermolecular attraction. Nonpolar molecules on 
the other hand are usually more volatile and soluble only in other nonpolar 

Dipole moments are given in electrostatic units or in Debye units, 
1 debye = 10~ 18 e.s.u. Values of this magnitude for e-l are realized 
since the charge of the electron is 4.8025 x 10- 10 e.s.u. and the distance 
/ between atoms is about 10~ 8 cm. Methods for obtaining dipole moments 
are given in physical chemistry texts. 






H 2 




H 2 S 




S0 2 




CH 3 OH 




NH a 


C0 2 




BCI a 


H 2 2 


SnCl 4 


QH 6 


CC1 4 


FIG. 3-15. (a) Random molecular orientation of polar molecules between uncharged 
condenser plates, (b) Condenser is charged and molecules line up in the field, partially 
neutralizing the plate charges and increasing the condenser's capacity as measured by 
the meter. The increase is related directly to the permanent and induced dipoles of 

the molecules. 


Dipole Moments and Molecular Structure 

The application of dipolc moment data is chiefly that of molecular 
structure elucidation. From what has been shown previously of hydrogen 
bromide (Fig. 3-13) and carbon dioxide (Fig. 3-10), one can surmise that 
HBr will have a dipole moment, since, due to the different electro- 
negativities of H and Br, the electrons in the bond are displaced toward 
the halogen. In CO 2 , although electronegativity differences exist the 
molecule is linear, so that the vectors of bond moments cancel and zero 
dipole moment is the result. Ammonia has a high dipole moment, indi- 
cating that it does not have a plane triangle shape as does boron trichloride, 
but instead is tripod shaped with the three hydrogens grouped toward 
one side (Fig. 3-11). 

The large dipolc moment of water shows that that molecule is angle or 
wedge shaped rather than linear. The use of two /; orbitals by oxygen 
should result in a 90 angle between the bonds, but it is actually about 
105 due to mutual repulsion of the H's. It can be shown that the ions 
of a salt are attracted to each other 1/80 as strongly in water as in air, 
hence in water they may be able to yield to the attraction of the H 2 O dipole 
and go into solution surrounded by solvent molecules. The ions are then 
said to be so/rated, and solution proceeds if the forces holding the ions in 
the crystal (lattice energy) arc less than the forces of attraction between 
solvent dipoles and ions (so/ration energy). The latter is directly related 
to the solvent's dielectric constant. The somewhat positive hydrogen 
side of the H 2 O molecule is attracted by negative ions and the negative 
oxygen side, with its unshared electron pairs is attracted by positive ions, 
thus making water a typical polar solvent and an excellent one for salts. 

A further result of the water dipole is its capability of promoting ioniza- 
tion (dissociation) of essentially covalent molecules like HBr and HC1. 
Bonds in these are partially ionic and break under the solvent's dipole 
influence. The resulting ions are then covered by a mantle of solvent 
molecules that hinders recombination and stabilizes the electrolytic 

Hydrogen Bonds 

The hydrides of the electronegative elements of the first period, NH 3 , 
H 2 O, and HF, have unusually high melting and boiling points when 
compared with hydrides of their own families. For instance, the boiling 
points of the hydrides of group VIA elements, H 2 O, H 2 S, H 2 Se, and H 2 Te, 
are respectively 100 C, 59.6 C, 42 C, and 1.8 C, and the question 
is: why should the hydride with the lowest molecular weight require the 
greatest amount of heat for vaporization ? This is answered by considering 


that an electrostatic attraction exists between the unshared electron pairs 
on an oxygen atom and the partially ionic hydrogens of an adjacent water 
molecule. Such bonding creates groups of molecules held together by a 
force not as strong as formal covalent or ionic bonds but stronger than 
van der Waals forces. Each water molecule is capable of forming four 
hydrogen bonds and grouping four other molecules about itself as two 
electron pairs and two hydrogens are available. Since these bonds are 
tetrahedrally distributed, one is, incidentally, able to account for the 
tetrahedral crystal structure of ice. The high dielectric constant for water, 
hydrogen fluoride, and hydrogen cyanide is also understandable in that 
hydrogen-bonded aggregates have a greater neutralizing ability on the 
experimentally applied electrical field than do the individual molecules 
derived from them. 

Density measurements on HF vapor show the gas to be composed of 
various associated molecules, (HF),. This and the fact that salts like 
KHF 2 are well known are evidence of hydrogen bonding in fluorine 
chemistry. A dotted line is used to designate this type bonding so the 
trimcr (HF) 3 would be H F - H F - - H F and the bffiuoride 
ion, (F - - H F) . 


1. Explain briefly by reference to the proper tables and figures: 

(a) Which element has the configuration Is 2 , 2s*2p*, 3 < s 2 3/> 4 ? 

(b) Why do the lanthanide series elements all have a valence of + 3? 

(r) What is the electronic arrangement of Ca and Ca+ 2 in the system used 
in part (a) above? 

(d) What is peculiar about the electron configuration for yttrium? 

(e) Why is the electroncgativity difference in HC1 greater than that in HI? 

2. J. P. Slipshod is a man who forms a hasty opinion from little data and 
has seldom been known to change his mind. Several of his observations follow. 
How can you answer each ? 

(a) "The idea of electrovalence will not work because when oppositely 
charged bodies like Na+ and Cl~ are brought together they discharge and the 
result is zero charge on both. 

(b) "The idea that a glowing piece of iron gives a continuous spectrum and 
a sodium vapor lamp a discontinuous spectrum is wrong. The iron glows red 
because its radiation is essentially all at the wavelength of red light. 

(c) "The silicon atom can be represented as l.v 2 , 2.y 2 2/A 3j 2 3/> 2 . The 3^ 
electrons are paired and the 3p unpaired. With only two unpaired electrons, 
the valence of Si can only be + 2, and whoever determined the formula of sand 
and quartz to be SiO 2 had better check his figures. 

(d) "Substances having the same electronic structure have the same size, 


since electrons in identical orbitals of different atoms have the same energy and 
are the same distance from the nucleus. Thus the experimental measurements 
that give the radii of Cl", A, and K+ (all of which have the electronic structure 
2, 8, 8) respectively as 1.81, 1.54, and 1.33 A are obviously wrong. Their values 
should be identical. 

(e) "The notion that ions can't have an independent charge greater than +3 
is bunk. Otherwise how does one explain Cr being -f6 in CrO^ 2 and Mn 
being -1-7 in MnO^, and how could one balance equations without valences? 

(/) "The atomic radii of B, C, N, O, and F are respectively given as 0.80, 
0.77, 0.74, 0.74, and 0.72. Anybody knows that can't be right since each 
succeeding element has one more electron and hence must be larger. This also 
holds for the supposed contraction in size in the rare earth series. 

(g) "Some people think that elements in a horizontal period show an increase 
in ionization potential with increasing atomic number because the electrons 
are added at about the same distance from the nucleus, but the nuclear charge 
is also increasing and holds ALL electrons more tenaciously. This notion is 
erroneous. Note for instance that B has a lower value than the preceding 
element, Be." 

3. From the electronegativity table: 

(a) Predict the stability of BF S , KBr, CI 4 , NI 3 , and PL,. 

(b) The heats of formation of several hydrogen halides are 

H 2 + Cl 2 = 2HCl + 44kcal; H 2 + Br 2 = 2HBr + 25 kcal ; H 2 + I 2 =2Hl + 3kcal 

Is this trend consistent with the table ? Explain. 

4. Write resonance forms for: (a) CO 2 showing the possibility of single, 
double, and triple bonds (b) CO^ showing why measurement of C O distances 
reveals that all are equal. 

5. According to the electron notation used in Fig. 3-10, show the bonding 
in sulfuric acid, (HO) 2 SO 2 , and point out the covalent and coordinate covalent 

6. From the table on dipole moments: 

(a) Argue for or against the proposal that hydrogen peroxide is a linear 
molecule with the structure H O O H. 

(b) Are the hydrogen halide values what one might expect by knowing the 
electronegativities of these atoms? Explain. 

(c) Can one determine whether the SnCl 4 structure is tetrahedral or square 
planar from the table data alone? 

(d) The boiling point of SnCl 4 is 1 14 C, and it is appreciably soluble in 
organic solvents. Is this behavior expected? 

(e) Two resonance forms for nitrous oxide are ~:N==N+ O: and 
:N^N + 6:~. Do these help explain why the measured dipole of N 2 O is 
approximately zero ? 

7. (a) Draw a series of simple diagrams showing how H 2 O dissolves NaCl. 
Liquid ammonia is also a solvent that dissolves many salts, giving electrolytic 
solutions. Explain how this is possible, and again picture the process. 


(b) Draw a picture of an ice crystal showing the arrangement of water mole- 
cules as described in the last section of the chapter. Postulate why it ought to 
be less dense than liquid water. 

8. CdBr 2 , but not NaBr, is appreciably soluble in organic solvents having 
low dielectric constants. What can one deduce about these salts? If one 
had dilute aqueous solutions of both, which would conduct electricity better? 

9. Why doesn't a rare earth series or long period exist in the first two hori- 
zontal rows of the periodic table? 

10. Give the number of electrons in each shell of Li+, B f3 , Be +2 , and C+ 4 . 
Arrange these ions in order of increasing size and explain. Arrange them in 
order of decreasing likelihood of independent existence. Explain. 

11. Magnetic moments studies show that Cr m and Co 12 have 3 unpaired 
electrons; V IV has one unpaired electron, and Mn m has 4 unpaired electrons. 
Show the structures of these ions (as illustrated in the examples given on p. 38) 
to account for the experimental measurements. 

12. From Table 3-4 plot three separate graphs of atomic number (abscissa) 
versus ionic radius; first for group I A ions, second for 11A ions, and third for 
univalcnt VIIA ions. Explain the trend in each plot. 

13. Using a compass and a scale of 1 A = 1 in., draw circles to represent 
the group I A and VHA monovalent ions. Show how Li I and NaCl might 
look as alternately packed spheres in a face of these salt crystals (the crystals 
have cubic symmetry). Repeat for KBr and CsF. 

14. Explain: 

(a) How ionization potentials can bemused to determine valences of elements. 

(b) The differences in electron structure between main group and subgroup 

(c) Whether a neon or an argon lamp (which depends upon ionization of the 
gas to give light) requires less voltage for starting, and why. 

W) On the basis of the size of the central element, why is H :J PO 4 a stronger 
acid than H : ,AsO 4 ? 
() Why is the dipole moment of O C -O zero but that of O S O is 1.6? 

15. (a) Plot a graph of electronegativity difference versus % ionic character 
from data given in the section on the electronegativity table. Extrapolate it 
to determine the ionic character of the bond in LiF. Predict from this some 
properties of lithium fluoride. 

(b) Using this graph, calculate the % ionic character of the compounds listed 
in problem 3(0). 

16. Show formation of H 2 O by the method of Fig. 3-11. Explain. 

17. (a) Fajans' rules have been stated in this chapter as conditions that lead 
to formation of ionic bonds. Restate them as conditions for co valence and 
briefly explain the reasons for each. 

(b) In the series LiF, Lid, LiBr, and Lil, the melting points are, respectively, 
870, 614, 547, and 446 C. Is this expected from what was stated in (a) concern- 
ing the effect of anion size on bond type? 

(c) In the series NaCl, MgCl 2 , A1CI 3 , and SiCl 4 , the respective melting points 


are 800, 708, 194 (4.2 atm.), and -70 C. Is this trend expected from what was 
given in (a) concerning the effect of ionic charge? 

18. Plot a graph of atomic number (abscissa) versus first ionization potentials 
of the elements from data in this book. Explain the positions of the group IA 
and group elements. 

19. A device for analyzing O 2 in flue gases does so by continuously measuring 
the magnetic susceptibility of the sample. This is possible because O 2 has two 
unpaired electrons. Make a suitable drawing to illustrate the molecule. 

20! (a) The boiling points of HF, HC1, HBr, and HI are 19.4, -85, -67, 
and 35.5 C, in that order. Explain the trend. 

(b) In the series He, Ne, A, Kr, Xe, and Rn, the boiling points show an 
increase in that order. Explain why one expects this trend rather than that 
shown in (a). 

21. Representing last level electrons with dots, show formation of these 
compounds from their atoms: MgF 2 , H 2 S, NC1 3 , LiBr, SO 3 , and SeS. Predict 
the bonding in each. 

22. (a) Predict which is more basic: Pb(OH) 2 or Pb(OH).,. Explain. 

(/>) Predict the trend in base strength in the series C(OH) 4 , Si(OH) 4 , Ge(OH) 4 
Sn(OH) 4 , and Pb(OH) 4 . Give the basis for your predictions. 

(c) Repeat (b) for the series Ce(OH) 3 , Eu(OH) 3 , Ho(OH) 3 , and Yb(OH) 3 . 

23. Some forms of the periodic table list general oxide formulas that are 
characteristic of vertical groups of elements. Fit these oxides to the proper 
groups: MO, M 2 O, MO 3 , M 2 O 7 , MO 2 , and M 2 O 3 . 

24. NO and NO 2 are paramagnetic but N 2 O and N 2 O 4 arc not. What can 
you deduce concerning their structures? 


1. J. A. Campbell, J. Chem. Educ., 23, 525 (1946). (Periodic table, atomic size) 

2. P. Bender, J. Chem. Educ., 23, 179 (1946). (Dipole moments) 

3. R. W. Taft and H. A. Sisler, J. Chem. Educ., 24, 175 (1947). (H bonding) 

4. W. F. Ehret, /. Chem. Educ., 25, 291 (1948). (Interatomic relationships) 

5. L. S. Foster, /. Chem. Educ., 29, 156 (1952). (Molecular shapes) 

6. A. A. Humffray, /. Chem. Educ., 30, 635 (1953). (Electronic theory) 

7. L. B. Clapp, J. Chem. Educ., 30, 530 (1953). (Covalent bond) 

8. R. T. Sanderson, J. Chem. Educ., 31, 2 (1954); 32, 140 (1955). (Electroncga- 

9. J. E. Clauson, J. Chem. Educ., 31, 550 (1954). (Periodic table) 

10. H. C. Longuet-Higgins, /. Chem. Educ., 34, 30 (1957). (Periodic table) 





In 1891 Alfred Werner, a German chemist, began studies on coordination 
compounds, the results of which were to open the large field of inorganic 
structural chemistry, yield important principles for understanding complex 
substances, and bring Werner a Nobef Prize in 1913. 

Early Experiments 

Some of Werner's first experiments were with cobalt salts. He found 
that cobaltous chloride in an aqueous mixture of ammonium chloride 
and ammonium hydroxide gave a dark solution that yielded three ammo- 
niated derivatives of air-oxidized tripositive cobalt. Chemical analysis 
showed them to have the molecular formulas of (a), (b), and (c) below: 

CoCl 2 + NH 4 OH -f NH 4 Cl-^CoCl 3 - 6NH 3 + CoCl 3 5N H 3 + CoCl 3 4NH 3 

() <//) (^ 

The products were peculiar in these respects: 

1. NH 3 could not be readily removed from any of them by reaction 
with strong base. Werner concluded ammonia was somehow tightly held 
since strong bases normally liberate weaker ones from their compounds. 

2. Upon electrolysis of their solutions, cobalt and ammonia migrated 
together to the negative pole while chloride went to the positive electrode. 
This indicated that NH 3 must be attached to Co in some form of positive ion. 

3. Compound (a) reacted with three equivalents of Ag f giving three 
equivalents of AgCl precipitate and another compound, (d), which was 



shown to be Co(NO 3 ) 3 -6NH 3 . Both (a) and (d) had molal electrical 
conductances corresponding to four ions, which was confirmed by freezing 
point measurements (see Chapter 5). Compound (a) should then pre- 
sumably be written [Co(NH 3 ) 6 ]Cl 3 to account for the observations. In 
solution it must exist as [Co(NH 3 ) 6 ]+ 3 + 3C1 -. Compound (d) should 
then be [Co(NH 3 ) r J(NO 3 ) 3 . 

4. Similarly, (b) was found to give three ions and (c) only two ions in 
solution. They were accordingly assigned the formulas [Co(NH 3 ) 5 Cl]Cl 2 
and [Co(NH 3 ) 4 Cl 2 ]Cl. 

Werner called these coordination compounds. Thousands of such 
compounds are now recognized. 

The Coordination Theory 

From such experimental observations Werner postulated his coordina- 
tion theory, which is now so well established as a unifying concept for 
understanding complex compounds that it is referred to as Werner's 
principle. The main points are : 

1. A number of metals, particularly among the transition elements, 
have two kinds of valence and they hold (coordinate) about themselves 
groups in two "spheres of attraction/* In the first sphere are those 
groups directly attached to the metal ion. In the second sphere are 
groups less strongly held. The central metal and its first sphere consti- 
tutes one ion in solution; the second sphere contains the remaining ions. 
Thus in the compound [Cu(NH 3 ) 4 ]SO 4 , the ammonia molecules are in the 
first sphere, the sulfate in the second, and, in solution, the ions 
[Cu(NH 3 ) 4 r 2 and SO^ 2 will be found. 

2. Each metal is capable of coordinating characteristic numbers of 
groups in the first sphere. Common coordination numbers are 2, 4, and 6. 

3. The central metal is capable of coordinating neutral groups such as 
NH 3 and H 2 O as well as negative groups like Cl~ 9 CN~, OR-, SCN~, 
NOg , F~, etc. The neutral groups do not contribute to the valence of 
the complex but the negative ions do; for the salt K[CrCl 4 (H 2 O) 2 ], 
K = + K 4C1- = 4, 2H 2 O = 0, so the oxidation state of Cr is (III) 
to give the molecule neutrality. 

4. Formation of complexes depends (as discussed later) upon the size, 
charge, and electronic distribution of atoms present. 

Nomenclature of Werner Complexes 

The following rules are a simplified summary of the International 
Union of Chemistry recommendations set forth in 1940. 

1. The cation is named first, then the anion. 

2. Coordinated groups which are negatively charged end in "o"; 


neutral groups have no characteristic ending. H 2 O is called aquo and 
NH 3 is ammine. 

3. The order of coordinated groups both in writing and naming is, 
negative groups, then neutral ones, as Na[CoCl 4 (NH 3 ) 2 ], sodium tetra- 

4. Prefixes denoting number are di, tri, tetra, penta, and hexa, respec- 
tively, meaning 2, 3, 4, 5, and 6. 

5. The central element's oxidation state is designated by a Roman 
numeral in parenthesis. If the oxidation state is zero, an Arabic (0) 
is used. 

(a) For complex cations, the number follows the element which is 
named in its ordinary way: [Cr(H 2 O) 6 ]Cl 3 is hexaaquochromium(HI) 
chloride, and [CoBr 2 (NH 3 ) 4 ]NO 3 is dibromotetramminecobalt(III) nitrate. 

(b) For complex unions* the numeral follows the name of the complex 
which ends in "ate": H 2 [PtCl e ], sometimes called chloroplatinic acid, is 
systematically named (di)hydrogen hexachloroplatinate(lV), and 
K 4 [Fe(CN) 6 ], known as potassium ferrocyanide, can also be called 
(tetra)potassium hexacyanoferrate(ll). In each of the last two systematic 
names, either term in parenthesis could be omitted and the names would 
still be unequivocal. Thus in the last, if 6 CN~ groups are coordinated 
with Fe 11 , it is evident 4 K p s are needed to complete the molecule. 

6. The prefixes bis, tris, and tetrakis are used before names of complex 
coordinated groups instead of di, tri, dnd tetra: [Co(en) 2 (NH 3 ) 2 ]+ 3 is the 
bis(ethylenediamine)diamminecobalt(III) ion. 

7. The symbol ft (mu) is used before the name of each bridging group in 
polynuclear complexes. The name of [(NH 3 ) 5 Ni O O Ni(NH 3 ) 5 ]SO 4 
is decaamine-//-peroxydinickel(II) sulfate. 

Some groups capable of being coordinated are: amido NH^, 
bromato BrO 3 , bromo Br~, carbonato --=- O 2 CO~ 2 , chloro Cl", 
cyano CN~, fluoro F, hydroxo OH~, imido =NH~ 2 , iodo I~, 
nitrato ONO 2 , nitrito ONQ-, nitro NO 2 , nitrosylo NO, oxalato 
O 2 C 2 O 2 , sulfato ^O 2 SO 2 2 , thiocyanato SCN~, and isothiocyanato 
NCS". Many complex organic groups are also known and may act 
as (chelating) organic reagents for reaction with and possible detection of 
the central metal. See Chapter 12. 

Complexes in the Qualitative Analysis Schemes 

The immediate purpose in developing some background in Werner 
chemistry is to enable one to understand and appreciate the role of complex 
formation in the laboratory procedures. The types listed below will be 

(1) Hydrated ions* or aquo complexes, are the rule in aqueous solution, 


and water may be held tightly enough to crystallize with the salt as water 
of hydration. Sometimes the hydrate formula gives the coordination 
number of the central ion; beryllium compounds, for example, are almost 
invariably tetrahydrated, suggesting that correct representation would be 
like [Be(H 2 O) 4 ]S0 4 . 

(2) Ammonia complexes are called ammines. Ag 1 , Cu 11 , Cd n , Ni 11 , 
Zn 11 , Co 11 , and Co 111 are ions forming ammines in the analysis described 
in later chapters. Separations of cations can be effected by means of 
complex formation. For example, insoluble Bi(OH) 3 forms with aqueous 
NH 3 , while copper and cadmium in the same solution give the soluble 
complexes [Cu(NH 3 ) 4 ]' 2 and [Cd(NH 3 ) 4 ]' a . The deep blue color of the 
tetramminecopper(ll) ion simultaneously identifies that metal in the 

(3) Sulfido complexes are numerous in the second cation scheme where 
Hg n , As 111 , As v , Sb lu , Sb v , Sn 11 , and Sn TV form soluble sulfides like 
AsSg 3 and HgS^ 2 in alkaline solution, while other group-2 metal ions do 
not. A separation based on this fact is used. 

(4) Hydroxo complexes are always possible in basic solution. The 
amphoteric cations of groups 2 and 3 form hydrated oxides that dissolve 
in excess strong base due to the formation of hydroxo ions such as 
Sb(OH)- Al(OH)j-, and Zn(OH)-' 2 . 

(5) Cyano complexes are the basis for a Cu Cd separation (see group 2) 
and for the identification by color of both ferrous and ferric iron (see 
group 3). Typical ions are Cd(CN) 4 2 and Fe(CN)~ 4 . 

(6) Hah complexes are present in halogen acid solutions of metallic 
ions, such as CuClj 2 , SbCl;: 3 , A1C1", etc. Except for fluoro complexes, 
they usually dissociate upon dilution with water. 

(7) Other anions capable of being coordinated and which are used in 
analysis include oxalato as Fe(C 2 O 4 )- 3 , thiocyanato as Fe(SCN)+ 2 , tartrato 
as Cu(C 4 H 4 O 6 )- 2 . and nitro as Co(NO 2 )~ 3 . 

Ionic Radii and Complex Formation 

The metal ions that best coordinate groups around themselves are those 
with a small radius and large charge. The ratio, charge/ tonic radius is some- 
times called the charge density. Stability of complexes is roughly propor- 
tional to this \alue, though other factors are also important. If this ratio 
is greater than approximately 2 (charge measured in electronic units and 
radius in Angstrom units. A) the possibility of complex formation is good. 

Sidgwick "The Effective Atomic Number" 

It seemed lotrical that valence characteristics of the central metal ion 
should be related to its coordination number but in Werner's time little 











No complexes 


-_*T| r 



Few complexes 

Cd 11 



Several complexes, ions hydrated 

Zn 11 



Generally complexed, ions hydrated 

Co u 



Generally complexed, ions hydrated 

Ni ir 



Generally complexed, ions hydrated 

Fe ru , Cr m 

__ -rjt 



Always complexed, ions hydrated 

Co 711 



Always complexed, ions hydrated 

Pt iv 



Always complexed, ions hydrated 

was known about chemical bonding, and his pure assumption of two 
bonding "spheres" was the most qualitative aspect of his theory. It is 
now known that the first sphere is formed largely by coordinate covalent 
bonding as each coordinated group is capable of electron pair donation; 
the second sphere is held by electro valence. 

The first working approach for correlating electronic configuration and 
coordination number is credited to N. V. Sidgwick (1926). Consider the 
cobalt atom and its tripositive ion for example. The atom (at. no. 27) 
has 2 electrons in the K level, 8 in the L, 15 in the M, and 2 in the N. The 
ion has 2 in the K, 8 in the L, and 14 in the M. If the ion were to coordinate 
6 : NH 3 , each ammonia has an unshared electron pair and would furnish a 
total of 12 electrons to the inner coordination sphere, enough to complete 
the 18 electron M orbit as well as re-establish and fill the N orbit with 
8 electrons. This gives the closed krypton configuration of 36 electrons, 
a number referred to by Sidgwick as the ion's effective atomic number or 
E.A.N. Fitting this idea are Zn 11 and Cu r which coordinate four groups 
to give the krypton structure, Fe 11 coordinating six groups to give the 
same structure, and Pd IV which takes six groups to add 12 electrons and 
give the xenon arrangement. Some exceptions are Cr ni , 6-coordination 
and E.A.N. = 33, Fe 111 , 6-coordination and E.A.N. = 35, and Ni li , 
6-coordination and E.A.N. = 38. These exceptions were never satisfac- 
torily dealt with by the Sidgwick theory, however, and it wasn't until 
after the development of quantum mechanics that a better explanation of 
the linkage in Werner ions was possible. 

Pauling The Quantum Mechanical Interpretation 

L. Pauling and others have extended the Sidgwick theory by means of 
magnetic susceptibility data and a quantum mechanical interpretation 



(Chapter 3) to give us our best ideas of bonding in transition metal 


Is 2s 2p 3s 3p 


4s 4p 

(1) Fe" 1 

(2) Fe 11 




same - 
- same - 

FIG. 4-1 . Ferrous ion (1) showing four unpaired electrons, followed by rearrangement 

(2) and complex formation (3) in which all electrons are paired. Dots represent iron 

electrons, x\s those contributed by six CN * groups. 

For example, FeCl 2 has a measured magnetic moment of 5.23 Bohr mag- 
netons, indicating four unpaired electrons. The compound K 4 [Fe(CN) 6 ], 
however, has no moment, indicating a pairing of all electrons. Ferrous ion, 
Fig. 4-1 (1), has two paired and four unpaired 3d electrons. If it is 
assumed that the unpaired ones pair up in the first three 3d orbitals, 
Fig. 4-1 (2), then six groups each carrying an available electron pair can 
coordinate with this metal ion. Quantum mechanics has shown that six 
equivalent bonds can be formed from the two 3d, the single 4s, and three 
4/7 orbitals, and further that the new bonds are directed spatially to the 
corners of a regular octahedron. X-ray evidence has proven this geometry 
in 6-coordinated complexes which are bonded by d*spP* bonds. The final 
structure of a cyano complex is seen to be the closed krypton configuration 
which is expected and found to be diamagnetic. 

Metals in the first transition series (sometimes called the iron group) 
have small energy differences among the 3d, 4$, and 4p orbitals (Fig. 3-5), 
and thus these orbitals can be combined in bonding. Transition metals in 
the second periodic row (the palladium group) use the 4d, 5,y, and 5p 
orbitals, and the transition elements of the third horizontal row (the 
platinum group) use combinations of the 5d, 6s 9 and 6p orbitals, which are 
also roughly energetically equivalent. The combination d 2 sp* always 
means an octahedral configuration. 

By the same methods, zinc ion (at. no. 30) can be shown to contain no 
unpaired electrons, its complexes with 4-coordination have the krypton 
arrangement and the compounds are diamagnetic. Available for holding 

* This means two d, one s and three p orbitals are hybridized giving six equivalent 
electron pair bonds. Table 3-4 gives the electron arrangements of atoms. 



the eight electrons donated by the coordinating groups are one 4s and three 
4p orbitals. The geometry of spP hybridization is tetrahedral. 


Is 2s 2p 



4s 4p 


FIG. 4-2. Zinc ion (1), and sp* bonding in a tetrahedral 4-coordination hydroxo 

complex (2). 

The situation in cupric compounds is a little different in that cupric ion 
has a magnetic moment associated with one unpaired electron as do its 
4-coordination complexes. X-ray examination indicates that cupric 
complexes are planar meaning that bonding other than .sp 3 , described 
above, is in effect. This has been shown to be dsp 2 as follows. Cupric 
ion has eight paired and one unpaired 3d electrons, Fig. 4-3 (1). In- 


Ls- 2s 2p 3s 3p 

3d 4s 


(1) Cu +2 

(T) (T) (TxTYT) (T) (TYiYT) 



(9\ fii 11 



\&) \AL 
(\\ p M Y"" 2 

onooo o 


\O) L/UA^ 

FIG. 4-3. Formation of a square planar cupric complex with dsp* bonds. 

stead of utilizing the 4s and three 4p orbitals, leaving the unpaired electron 
in the 3d state and giving the tetrahedral sp? bond type, the odd electron is 
promoted from its original state to a 4p orbital as in Fig. 4-3 (2). The 
eight electrons of the incoming 4X:~ groups occupy the vacant 3d, 4s 9 
and two 4p orbitals. Bonds of the type dsp 2 give a square planar structure. 
The observed magnetic moment for [Cu(NH 3 ) 4 ]SO 4 is 1.82 and that 
calculated for the compound on the basis of one unpaired electron is 
1.73 Bohr magnetons. 

There are many combinations of orbitals known, resulting in various 
numbers of bonds and geometrical distributions of groups around a 


central atom. The bonds which usually form are those having minimum 
energy (greatest strength), and not infrequently these are hybrid bonds. 
In addition, the size of coordinating groups and of the central metal is 
also a factor, maximum coordination number being favored with small 
groups. Maximum stability of complexes is found within certain ratios 
of sizes of the atoms present. These points are summarized for several 
cases in the following table. 



Central Metal 






MX 2 Ag 1 , Hg n , Au 1 





MX 3 Ni 11 , Cu 11 



Triangle planar 

: 0.15 

MX 4 

Cu 1 ', Zn, 11 Ni 11 , 
Ag n ,Au m , Pd 11 , 

p t !I 


dsp 2 

Square planar 

- 0.41 

MX 4 

Cu 1 , Ni 11 , Al m , 




'- 0.22 

Zn 11 , Cd 11 , Hg 11 , 

Sn lv , As v 

MX 6 

Co 111 , Co'MMi 11 , 




".- 0.41 

Fe lir ,Fc ll ,Mn ni , 

Sn IV ,Al lll ,Sb 111 , 

As 111 

MX 8 

Os lv ,W lv ,Mo lv , 


" vt 


-.- 0.65 

* M stands for the central metal ion and X for the coordinating group. In the 
column below, the symbol - is read "equal or greater than." 
t Of the nine orbitals, only eight are used for bonding. 

The most frequently encountered types from Table 4-2 are: 
I. Linear ^-coordination is illustrated by [H 3 N:-> Ag-:NHJ f in 
group 1 cation analysis and its formation accounts for the solubility 
of AgCl in NH 4 OH. In like manner, when sodium thiosulfate ("hypo") 
is used in photography to wash unreactcd silver halide from film, 
[O 3 SS:-> Ag^-rSSOJ- 3 is the soluble complex formed. In cyanide 
plating baths or the process for dissolving gold and silver in cyanide ore 



treatment [N^=C:-> Ag^-:C=N]- and [N C:-> Au-:C~N]- are 
products. In these formulas, arrows represent coordinate covalent bonds. 

2. Planar ^-coordination is illustrated below in the tetraiodomercury(II) 
ion. See test 19-9 (b) for a preparation of K 2 [HgI 4 ]. 

3. Tetrahedral ^-coordination is pictured below for the hydrated zinc 
ion. The sp 3 bonds are directional to the corners of a regular tetrahedron. 



FIG. 4-4. (a) Two representations of a planar ion. (h) A tetrahedral ion. Dashed 
lines show the symmetry of the figures; arrows show coordinate covalent bonds. 

4. Octahedral ^-coordination can be shown in several ways as indicated 
in the figure. 


H 3 N:< 

FIG. 4-5. Two ways to show the geometry of an octahedral ion. Designation of 
bond types is omitted for simplification. 

Isomerism in Werner Ions 

One of the convincing features of the Werner theory was its ability to 
predict the possibility of isomers, just as the electron orbital idea was 
later able to predict the geometry of ions from a knowledge of the bonding 
orbitals that were used and their spatial orientation, homers are sub- 
stances with the same molecular formula but different arrangements of 
atoms. Isomerism was well known in organic chemistry (CH 3 CH 2 OH 



= ethyl alochol, and CH 3 O CH 3 = dimethyl ether, two distinctly 
different, stable, C 2 H 6 O compounds), but Werner was the first to apply the 
same thinking to inorganic chemistry. 

1. lonization isomerism refers to isomers that give different ions in 
solution because of their inclusion in either the tightly held coordination 
sphere or the readily dissociated second sphere. Two such isomers are 
[CoCl 2 (NH 3 ) 4 ]Br and [CoClBr(NH 3 ) 4 ]Cl. The first will give yellowish 
AgBr with Ag + in solution, whereas the second will give white AgCl. 

2. Manner-of -attachment (salt) isomerism is the result of isomeric forms 
of certain polyatomic ions coordinated with the metal in different ways. 
Thus nitrite (a) and nitro (b) are isomeric and will give different series of 

(a) _O N-= 




These happen to be interconvertible with the nitro group, the more stable 
form. "Sodium cobaltinitrite" as prepared in Test 19-7 (b) is a nitro 

H 3 N- 




(a) cis 


(b) trans 


FIG. 4-6. In this example the square planar structure can have two forms and the 

tetrahedral only one. Since dichlorodiammineplatinum(Il) is known in two forms, its 

geometry is planar. The bonding in Pt 11 complexes is dsp 2 , which is calculated to be 

planar. Structure (r) is therefore incorrect for the ion. 

FIG. 4-7. An example of cis-trans isomerism. The chlorines are the reference atoms. 
The solid lines are used to show the symmetry of the ions. 



3. Geometrical isomerism is also called cis-trans isomerism. It can 
occur with octahedral 6-coordination (complexes like MX 4 Y 2 ) and planar 
4-coordination (complexes like MX 2 Y 2 ). The as isomer has two refer- 
ence groups adjacent to each other, the trans form has the groups 

Chelate Compounds 

Any complexing group capable of being a donor of two or more electron 
pairs is called a chelating agent, and the substance formed is called a 
chelate. The attachment is again by co- 
ordinate covalent bonds and the atoms 
best known for electron sharing in chela- 
tion are oxygen, nitrogen, and sulfur. If 
a chelating group attaches in two positions 
in the coordination sphere it is called a 
bidentate group (literally, having two teeth). 
Tridentate, tetradentate, pentadentate, and 
hexadentate groups are also known. Ring 
formation is a characteristic of chelation, 
and the bond angles of common atoms 
like C, O, N, and S in chelating groups 
favor 5- and 6-membered rings. The pre- 
sence of chelates is detected by (a) isola- 
tion and analysis of pure compounds 
(/?) changes in color (c) low conductivity, 
and (d) low solubility in water but high 
solubility in organic solvents. Chelates 
are useful in analytical chemistry (see 

Chapter 12 on organic reagents), in softening water, washing up areas 
contaminated with radioactive salts, etc. Natural chelates such as 
chlorophyll and hemoglobin are also known. 

A compound of recent practical interest is ethylenediaminetetracetic 

O O 

II .. .. II 

(HO C CH 2 )r-N CH, CH 8 N (CH f C OH) 2 . 

It powerfully complexes divalent metals and is used in titrating Ca f2 and 
Mg^ 2 in water analysis, in dissolving scale from boilers, etc. It can be 
tetradentate or hexadentate; the latter is illustrated. 


FIG. 4-8. Hexadentate chela- 
tion of Versene with Ca ' 2 . The 
dashed lines are used to aid in 
showing the octahedral configura- 

* Two trade names are Versene and Sequestrene. Several derivatives of the acid 
have also received attention. 


Stabilization of Oxidation States 

Complex formation can stabilize oxidation states of metallic elements 
that normally cannot exist in aqueous solution. For instance Co m and 
Co lv are powerful enough oxidants to liberate O 2 from H 2 O, but their 
ammine derivatives are stable in water solution. The chemistry of such 
unfamiliar oxidation states as Cu IH , Ag 11 , Ag llr , Cr 11 , Mn 1 , etc., has been 
conveniently studied in the presence of complexing agents. 

Lattice Compounds 

Werner compounds involve definite chemical bonding between the 
central ion and the groups in the coordination sphere, and the resulting 
complex ion is largely retained in solution. There is a general class of 
substances known as lattice compounds which from their composition 
might be taken for coordination compounds but which should be catalogued 
otherwise. Most inorganic lattice compounds are double salts, so called 
because two salts (plus water frequently) crystallize together in definite 
molecular ratios to form a stable, symmetrical crystal lattice. Such 
compounds disappear upon dissolution, however, and all possible ions 
are free to give their characteristic reactions. 

Other Complex Inorganic Compounds 

There are many complex inorganic substances, but by comparison to 
large molecular weight organic compounds they are less numerous. Three 
examples are given below. 

1. Polynuclear Werner salts are those containing two or more metallic 
ions with some coordinated atoms held in common. Basic salts generally 
belong to this class; basic lead carbonate, PbCO 3 2Pb(OH) 2 , for example, 
is better represented as 

H H 

O x Ov 



H H 



2. Poly acids are formed by combination of acid anhydride molecules 
such as V 2 O 5 , MoO 3 , and WO 3 with another acid that furnishes a central 
ion for the establishment of a complex anion. If only one kind of 
anhydride molecule is present, the compound is called an isopoly acid; if 
more than one, it is a heteropoly acid. These substances are quite complex 
and capable of wide modification under different experimental conditions. 
See Chapter 20 and test 22-16 for some specific examples. 

3. Silicates can be thought of as polymers in which oxygen atoms 
are coordinated octahedrally around dipositive cations like Mg+ 2 or 


tetrahedrally around Si IV or Al 111 . These are of interest because of 
the prevalence of silicate minerals. 


1. (Library) There is a generalization which states that in many Werner ions 
the coordination number is twice the central element's oxidation state. From 
a reference book, list a number of examples both supporting and contradicting 
this, and give your conclusions. 

2. The ionic radii of several metallic ions are V m = 0.66, Ti lv = 0.68, 
Be ir = 0.31, Ba 11 = 1.35, and Rb l = 1.69 A. Calculate the charge densities 
and predict whether or not each is prone to complex formation and hydration 
in solution. How many groups will probably coordinate with each? 

3. [Co(NH 3 ) 6 p is more stable than [Co(NH 3 ) 4 ]+ 2 (the latter liberates H 2 
from H 2 O). Explain the stability difference by both the Sidgwick and Pauling 
methods. (For the latter, the odd electron of Co IT is promoted to the 4d state.) 

4. Dipositive nickel may under different conditions give either square planar 
or tetrahedral 4-coordination. As in Fig. 4-1, show the electronic distribution 
and the orbitals used in bonding from the following hints: (a) assume that in 
the planar configuration, two unpaired 3d electrons of Ni+ 2 pair up leaving one 
open 3r/ orbital, and (b) whereas the planar ion is diamagnetic, the tetrahedral 
ion is paramagnetic to the extent of indicating two unpaired electrons. 

5. (a) (Library) In a dictionary find 'the definition of the word chela and 
relate to the action of chelating groups. 

(b) (Library) Look up the definition of sequester and explain why Sequestrene 
is a descriptive trade name for ethylenediaminetetraacetic acid. 

(c) A sodium salt of ethylenediaminetetraacetic acid was suggested once as 
an internal medicine for dissolving gall stones. Explain. (It required too 
high a /?H to be valuable, however.) 

6. There is a generalization that says that those metallic ions already having 
a rare gas configuration tend not to form Werner ions. Cite several specific 

7. [W(CN) 8 ]~ 4 is a stable complex ion; explain this by means of the effective 
atomic number idea. 

8. Four octahedrally coordinated isomers of CrQ 3 - 6H 2 O are known. Draw 
structures from these descriptions: (a) gives 4 ions in solution (b) gives 3 ions 
(c) gives 2 ions and has a cis configuration, and (d) gives 2 ions and has a trans 

9. Interpret the following: (a) CoCl 3 -5NH 3 -H 2 O is not dehydrated at 30 C 
over a drying agent but from its solution, Ag+ precipitates three equivalents of 

(b) A certain compound is known to contain platinum, ammonia, and chlorine. 
When 125.0 mg of it are boiled with a solution of strong alkali, the compound 
slowly decomposes. The liberated ammonia is absorbed in boric acid and 


requires 11.25 ml of 0.1097 N HC1 to titrate it (see Ammonium, Chapter 19). 
The residue of platinum in the main reaction flask is washed, filtered off, dried, 
and found to weigh 60.3 mg. In a separate experiment, a 10~ 3 //; solution of 
the compound has a molal conductance of 265 (see Table 5-3). Show that 
correct interpretation of this data leads to the conclusion the compound is 
dichlorotetrammineplatinum(IV) chloride. Write the structural formula. 

10. Name the Werner compounds: Na[Sb(OH) 6 ], (NH 4 ) 2 [Sn(OH) 6 ], 
K 2 Na[Fe(CN) 6 ], [A1(OH) 3 (H 2 O) 3 ], and [NiCl 3 (NH 3 ) 3 ]Br. 

11. (Library) Find the structure of (a) hemin and (b) chlorophyll. Copy them 
and explain why they are chelates. (r) Find reference to clathrate compounds. 
Define the term and give an example. 

12. (Library) Find reference to the following types of water of crystallization: 
(a) constitutional (b) coordinated (c) anion (d) lattice, and (e) zeolitic. Define 
the terms and give an example of each. The listing here is in the order of 
decreasing stability. Can you offer any reasons for this order? 

13. How could one chemically distinguish among potassium chromium alum, 
hexamminechromium(III) sulfate, and potassium disulfatodiamminechromate 

14. Predict which member of the following pairs will be more stable and tell 
why: (a) Na 2 [HgQ 4 ] or Na 2 [HgI 4 ] (b) Na 4 [NiQ 6 ] or Na 3 [NiCl ]. 

15. Show electronically that [Co(NH 3 ) ]+ 3 is expected to be octahedral 
because of d 2 xp 3 bonding. 

O O 


16. The oxalato group O C C O is bidentate. Draw the octahedral 
trioxaltoiron(lll) ion. Draw the isomers of dichlorodioxalatoiron(III) ion. 
Explain why a radiator cleaner sold for rust removal contains oxalate. 

17. Converted to standard temperature and pressure, it is found that 56 cc 
of ferric chloride vapor weighs 811 mg. Calculate the mol. wt and postulate 
a structure for the compound. 

18. (a) [Cd(NH 3 ) 4 ]+ 2 has no magnetic moment. Give the electronic con- 
figuration and draw the ion with its correct geometry. Explain your choices. 

(b) [Cr(NH 3 ) 6 ]^ 3 has a moment associated with three unpaired electrons and 
the complex is octahedral. Give the electronic structure, and indicate the 
orbitals used in bonding. 

19. A solution contains palladium, ammonia, and chloride. Analysis shows 
that there is 0. 1 g atoms of Pd per 100 ml and that the ratio of Pd to Cl is 1 to 4. 
Addition of 100ml of 0.4 F AgNO 3 to 10ml of palladium solution gives a 
precipitate of 20 mmoles of AgCl. Addition of 10 ml of 0.2 F HNO 3 to another 
10 ml portion of sample gives a solution 0.1 F in H + . What can one deduce 
concerning the original solution ? 

20. When 3.27 g of the compound NaF-CoF 5 are dissolved in 100 g of H 2 O, 
the solution's freezing point is found to be 0.60 C. In an electrolysis of the 
solution at room temperature, the cobalt concentration (as measured visually 
by depth of color) increases around the positive electrode. What can one 



1. D. T. Hurd, J. Chem. Educ., 25, 394 (1948). (Complex ions) 

2. R. L. Pecsok, /. Chem. Educ., 29, 597 (1952). (Versene complexes) 

3. A. E. Martell, /. Chem. Educ., 29, 270 (1952). (Complexes in aqueous solution) 

4. D. H. Busch, /. Chem. Educ., 33, 498 (1956). (Coordinate bonds in complexes) 

5. Z. G. Szabo and M. T. Beck, Anal. Chem. 25, 103 (1953) (Complexes in 

6. A. Werner, New Ideas on Inorganic Chemistry, English trans., Longmans, Green, 
New York (1911) 

7. A. E. Martell and M. Calvin, Chemistry of the Metal Chelate Compounds, Prentice- 
Hall, New York (1952) 

8. T. Moeller, Inorganic Chemistry, John Wiley & Sons, New York (1952) 






Early Theories of Acids and Bases 

The theories developed to explain and correlate the behavior of acids 
and bases date back to the alchemists. Those early chemists recognized 
such substances as chemical opposites in that interaction caused a mutual 
loss of characteristic properties. They also noted that acids tasted sour, 
changed the colors of plant extracts (the first indicators), etched metals, 
dissolved some minerals, etc., but, since little was known about chemical 
composition, it was not until the early part of the nineteenth century that 
quantitative statements could be made. 

Before 1814 it was believed that all acids contained oxygen, but in that 
year the English electrochemist H. Davy provided a method for compound 
decomposition with new experimental techniques. He showed that 
electrolysis of HC1 solution gave only H 2 and C1 2 , and from further experi- 
ments on H 2 SO 4 and HNO 3 guessed that, since hydrogen was the only 
element common to all the acids, it might be responsible for their acidic 

In 1835 the German chemist J. Liebig conducted a number of tests to 
show acid action on metals and concluded that since H 2 was released, and 
compounds were formed that contained the metal, acids should be defined 
as agents containing hydrogen replaceable by metals. In a continuation 
of electrical experiments in the same year by M. Faraday, that famous 
English chemist found that most acids and bases give electrically conducting 



solutions that can be decomposed by direct current. He coined the 
word electrolyte for substances yielding conducting solutions. 

In the following years electrolytes and electrochemical phenomena were 
extensively investigated and it soon became apparent that two kinds of 
electrolytes existed. There were those that showed weak conductivity 
and those that showed strong conductivity. Furthermore, the weak 
electrolytes were noted to exhibit a gradual increase in conductivity with 
dilution, whereas the strong electrolytes demonstrated excellent conduc- 
tance at almost any dilution. 


It wasn't until the period 1883-1887, during which S. Arrhenius 
developed his theory of electrolytic dissociation, that a unifying idea was 
able to explain most of the observations. The Swedish chemist proposed 
that in water solutions, strong electrolytes (strong acids and bases and 
most salts) exist largely as charged particles called ions rather than as 
molecules. These free moving ions are capable of carrying current, and 
move toward the oppositely charged electrodes during electrolysis. His 
conclusions were based partly upon experiments he had made of the 
freezing points of solutions of electrolytes in which it was noticed, for 
example, that an HC1 solution gave about twice the expected freezing 
point lowering and Na 2 SO 4 about three times the expected effect. This 
meant that more particles were in solution than was obvious from the 
molecular formulas. 

Arrhenius therefore defined acids as substances of the type HX dis- 
sociating in aqueous solution to yield H+ and X~, bases like MOH as 
substances giving M + and OH~, and neutralization as the result of these 
ions combining: 

HX = H^ + X- 

MOH = M+ + OH- 
H+ + OH- = H 2 

He visualized that these ions would be separated by a polar solvent 
like H 2 O, since coulombic forces between ions is low and further, that 
increasing dilution should encourage ionization since the ions would 
become more independent. This concept satisfied all observations on 
weak electrolytes admirably and is still held today. He showed by 
electrical conductivity methods (special experiment 1 and Table 23-1) 
that the conductivity of an equivalent weight of a weak electrolyte increased 
gradually with dilution and reached a maximum value in very dilute 
solution, at which point presumably all molecules of solute had dissociated 
into ions. 


When Arrhenius applied the same experimental methods to strong 
electrolytes he ran into trouble, however. For instance he found from 
conductivity and freezing point data that 0.1 M HC1 is apparently 89% 
dissociated, 0.01 M is 96%, and 0.001 M is 98%. He never did explain 
the basis for the differences in the two types of electrolytes and could not 
therefore answer the question: why aren't strong electrolytes always 
100% dissociated? The dilemma was finally solved 35 years later by 
Debye and Huckel, but this does not detract from the insight and original 
thought he gave the problems. Briefly, the Arrhenius contributions are 

(1) Explained why compounds in solution conduct electricity. 

(2) Gave consideration to the role of polar and nonpolar solvents in 
aiding or hindering dissociation. 

(3) Explained the results of electrolysis experiments: ions migrate to 
electrodes of opposite sign. 

(4) Permitted characterization of compounds on the basis of the 
maximum conductance their ions give at a dilution when further addition 
of water gives no further dissociation ("infinite dilution'). 

(5) Formulated an explanation of why solutions of strong electrolytes 
give abnormally large changes in boiling and freezing points and osmotic 
pressures (since these colligative properties depend upon the number of 
particles in solution). 

(6) Showed that compounds that can give several ions in solutions will, 
in the case of weak electrolytes, dissociate in steps: 

H 3 P0 4 = H^ + H 2 POj (5-1) 

H 2 PO 4 = H+ + HPO~ 2 etc. 

This also accounted for neutralization of one or more H + by stepwisc 
titration with base and for the formation of salts such as NaH 2 PO 4 , etc. 

(7) Explained why solutions containing a given ion all have reactions 
characteristic of that ion. Thus, both Nad and HC1 solutions give a 
precipitate of AgCl with Ag+, but HC1O 4 does not since its dissociation, 
HC1O 4 = H+ + ClOj, gives no Cl~. CC1 4 also gives no reaction with 
Ag+ because it is a nonelectrolyte. Carrying the last illustration a step 
further, if CC1 4 is a nonelectrolyte, then chloroform CHC1 3 is by analogy 
expected to be a nonelectrolyte and hence not acidic, which is the case. 

(8) Formed a foundation for the treatment of weak electrolytes by mass 
action law (Chapters 6 and 7) by picturing them in solution as a mixture 
of undissociated molecules and ions, the latter increasing as the solution 
is diluted. Acetic acid as a typical compound is therefore correctly 
described by, HAc ^ H+ + ^ (5 _ 2) 

Molecules Ions 


Recognition of Interionic Forces 

Strong electrolyte interpretation was reconsidered by three investigators 
in the period 1900-1912. S. Jahn surmised that since ions in solution are 
charged particles, unlike charges will attract each other and perhaps 
decrease the over-all conductance efficiency. 

G. N. Lewis showed that because of interionic attractions and repulsions, 
corrected concentrations should be used in calculations. He called these 
effective concentrations, or activities, and their use is the basis for refined 
treatment today. 

G. B. B. M. Sutherland, in similar work, attempted mathematically to 
define three forces at work in solutions that complicate affairs by tending 
to immobilize ions: (a) ion-solvent attraction (b) ion-ion attraction and 
(c) solution viscosity. His results were only semiquantitative, however. 

Although none of these men gave a complete explanation of the charac- 
teristics of electrolytic solutions, their contributions greatly aided their 
successors and some parts of their thinking are still retained. 

Debye and Hiickel 

As will be explained in Chapter 13, the development of X-ray diffraction 
in the decade 1912-1922, which made structural chemistry an exact 
science, gave chemists the data needed for continuation of many problems 
P. J. W. Debye and E. Hiickel (1923) began with the knowledge from 
X-ray study on salts that both solid and gaseous NaCl is composed 
exclusively of ions Na+ and Cl~ and is to be considered 100% ionic at all 
times. Whereas Arrhenius had thought the situation to be one of 
equilibrium between undissociated molecules and ions (which is true for 
weak electrolytes), as NaCl = Na 1 " + Cl~, Debye and Hiickel proved 
that in solution the picture is better given by 

= Na f + CI- (5-3) 

The left side specie is sometimes called "bound ions" to signify decreased 
mobility by interionic attraction, and the right side, "free ions,' 9 meaning 
those far enough apart to behave as individuals. 

Bronsted and Lowry 

With weak electrolytes explained by Arrhenius, strong ones by Debye 
and Hiickel, and in the meantime considerable information on chemical 
bonding having become known, attention was turned to refining acid-base 

J. N. Bronsted and also T. M. Lowry (1923) recognized limitations in 
defining acids and bases simply in terms of H+ and OH- because closely 
related phenomena were not correlated. For example, although H 3 PO 4 


was called an acid, its ion, H 2 PO7, which also could give H+, was not 
always classed as an acid, nor was it classed as a base when it combined 
with H+ to reform H 3 PO 4 . Accordingly they defined an acid as any 
substance which can donate a proton (W ), and a base as any substance 
that can accept a proton : 

Acid = H+ + base 

HC1 = H+ + Cl- (5-4) 

H 2 PO^ = H+ + HPOj 2 etc. 

In an aqueous solution of an acid the base is H 2 O, whose unshared 
electron pairs make it capable of holding the proton. The typical 
system is an equilibrium in which two bases compete to the extent of their 
base strengths for protons from the two acids present:* 

HC1 + H 2 = H 3 0^ + Cl- (5-5) 

Acid 1 + base 2 = Acid 2 + base 1 

In Bronsted nomenclature, acid 1 and base 1, and acid 2 and base 2 
respectively constitute two conjugate pairs, the members of each pair 
differing only by an H H . Acids may be cations like H 3 O+, anions like 
HSO~, and molecules like HAc in this system, since any of these can give 
protons. Substances like HSO 4 may be acidic in a basic environment or 
basic in an acidic environment and are said to be amphiprotic or are called 

Acid-base strengths^ are capable of comparison by the Bronsted theory. 
In this matter the role of water as a proton donor or acceptor is of prime 
importance. With a base stronger than water itself, H 2 O dissociates 
and acts as an acid : 

S- 2 + H 2 = HS- + OH- (5-6) 

JJl A2 Al Ji2 

With an acid stronger than itself, water is capable of proton acceptance: 

+ H 2 O = H 3 O+ + SOj 2 (5-7) 

Al 1V2 Ai> BL 

By measuring pH (Chapter 6) of a series of reactions like the two above, 
one can establish a list of approximate acid and base strengths. 
Strong acids and bases cannot be evaluated in this manner, however, 

* This means that Bronsted calls the acid in aqueous solution H 3 O f the hydronium 
ion. Though this will be inferred in following discussions, H f will be used instead of 
H 3 O+ in most cases. The simplification is made because the extent of proton hydration 
varies and because certain conventional expressions like />H (instead of /?H 3 O) have 
been retained. Where H 3 O f is vital to the text, it is shown. 

t Quantitative aspects are considered in Chapter 7. 


since they all appear equally strong. It is then necessary to resort to 
measurements in solvents with lower acid strength for the examination of 
strong bases and with lower base strength for the investigation of strong 
acids to obtain a spread in proton transfer ability. Acetic acid is a 
solvent of the latter type. It has acidic properties of its own of course 
but is forced into the role of a base when confronted with stronger proton 

:0: :0: 

HC1 + CH 3 Of = CH 3 C/ + Cl- (5-8) 


A I B2 " A2 HI 

Acid strength of Al (dry HC1 gas) in this solution is measured by electrical 
conductivity, since strong acids will shift the system to the right with the 
production of ions. Experiments show the strength order of several 
common acids is HC1O 4 > HI > HBr > HC1 > HNO 3 , though in H 2 O 
they appear equally strong. 


E. C. Franklins research in the United States during the first quarter of 
this century began by considering that in water a solvent-solvent reaction 
is possible: 

H 2 O + H 2 O = H 3 O' + OH (5-9) 

Al H2 A2 III 

This idea should carry over into other solvent systems as liquid ammonia 
and anhydrous acetic acid. For example: 

NH 3 + NH 3 = NH+ + NH 2 (5-10) 

Al 112 A 2 HI 

HAc + HAc = H 2 Ac H + Ac~ etc. 

A I H2 A2 HI 

He successfully developed a system of ammonia chemistry which clearly 
showed the value of general acid-base theory for understanding a great 
variety of reactions. For instance in liquid ammonia, the reaction between 
sodium amide and ammonium chloride (the reverse of eq. 5-10) is thought 
of as a neutralization, analogous to NaOH and HC1 reacting in water, 
since a solvent and a salt is the result: 

NaNH 2 + NH 4 C1 = NaCl + 2NH 3 (5-11) 


Between the years 1915 and 1938, the American chemist G. N. Lewis 
carried on acid-base studies and from these discovered many examples in 
support of four criteria oj acids and bases on which his work dwelled: 


(1) Neutralization reactions are rapid. 

(2) Stronger acids and bases will replace weaker acids and bases from 
their compounds. 

(3) Titrations using indicators to find equivalence points are feasible. 

(4) Both acids and bases frequently function as catalysts. 

The Lewis concepts are more general than those of Bronsted because H+ 
is not invariably necessary. Instead, Lewis considered the mode of bond 
formation as the characteristic feature of acid-base interaction, and the 
presence or absence of unshared electron pairs as the distinguishing 
feature of bases and acids themselves. He defined an odd as any substance 
that can accept an electron pair (frequently it has only six of its normal 
octet of valence electrons), and a base as any substance capable of donating 
the electron pair (it has an octet of electrons but not all are used in bonding). 
Reaction between the two results in the formation of a coordinate covalently 
bonded product (Chapter 3). This view incorporates previous examples 
as well as others, including reactions in the gas and solid phases, and of 
many organic compounds. A few examples follow. 

H+ + :OH- = H<-:OH (5-12) 

Acid base Neutralization product 

(needs f pair) (has unshared ) (has coordinate covalent bond) 

BF 3 + :NH 3 = F 3 B-:NH 3 (5-13) 

Si0 2 + MgO = MgSi0 3 (5-14) 


The latest (1938) and most inclusive acid-base ideas are due to the 
Russian chemist, M. I. Usanovich. He has defined an acid as any substance 
capable of (a) yielding a cation (h) combining with electrons or an anion, 
or (c) giving a salt when it neutralizes a base. A base is any substance 
which can (a) give electrons or onions (b) combine with a cation, or (c) 
neutralize an acid to yield a salt. This incorporates all older theories and 
extends them again, but because it is so general, the Bronsted and Lewis 
views still are used for most acid-base interpretations. 

Electrical Conductance of Electrolytes 

As has been inferred, electrical conductance studies did much to not only 
advance acid-base ideas but our general understanding of all electrolytic 
solutions. Since ions transfer current through a solution, their number 
will be directly proportional to the conductance (and vice versa) under 
specified conditions. (See special experiment 1 for typical procedure 
and experimental data.) 



The specific resistance, p (rho), of a solution is the ohm resistance 
measured between parallel faces of a cube of solution 1 cm on an edge. 
The reciprocal of the specific resistance or resistivity is the specific conduc- 
tance, kappa, or K = 1/p. Conductance is expressed in reciprocal 
ohms (ohms* l ) which are also called mhos. If the temperature is constant, 
the conductance of a solution will vary with concentration and the speed 


0.2 0.3 4 

Concentration (N) 



FIG. 5-1. Equivalent conductance vs. concentration for a typical weak and strong 


with which the ions carry the current, since different ions have different 
mobilities. As one might suspect, H f and OH have greater speed or 
mobility in water solution than do other ions. To study mobility, a 
standard quantity of solute must be defined for reference and the gram 
equivalent weight is used. The conductance of this quantity is called the 
equivalent conductance at the particular dilution, and abbreviated with a 
capital lambda, A. To get A, one measures the conductance of 1 cc of 
solution then multiplies it by K, the cc volume needed to contain the 
gram equivalent weight, or A = K*. 

Table 5-1 shows that the equivalent conductance of both strong electro- 
lytes (NaCl, K 2 SO 4 , KOH, HC1) and weak electrolytes (NH 4 OH, HAc) 
increases with dilution, but only values for the former are high at all 
concentrations. Both types of electrolytes (in more dilute solution) 
attain a maximum value, after which, further dilution has no effect on A. 
The equivalent conductance at this "infinite dilution' 9 or "zero concentra- 
tion" is abbreviated A . At that concentration all solute ions function 
unencumbered by interionic repulsions or attractions. 



In practice one measures conductance of a strong electrolyte at a 
number of reasonable dilutions, then plots A versus the normal concentra- 
tion N, and extrapolates to zero concentration. This is not feasible for 





K 2 S0 4 


NH 4 OH 




10 3 








10 4 
















10 6 







weak electrolytes since the graphical extrapolation is too uncertain and 
the method described below is used. 

The Law of Independent Ion Migration 

From conductance studies carried out in the years 1870-1890, the 
German chemist F. Kohlrausch assumed that the equivalent conductance 
of any electrolyte at "zero concentration" is the sum of the zero concentra- 
tion (or limiting) equivalent conductance of its cation A+ and anion h~. 
These ionic mobilities are independent of each other and can be added 
directly to obtain A for the compound. Ionic mobilities are related to 
the speed with which ions move under an applied voltage and are calculated 
(in physical chemistry courses) from A and transport numbers, which are 
the fractions of current carried by each ion. 

AT 18 AND 25 C 


A+, 18 C 

#, 25 C 


AjT, 18 C 

A ~, 25 C 










*so r * 









iBa+ 2 






Ka +2 



ic 2 o 4 - 2 
















Calculation of Dissociation by the Conductivity Method 

The equivalent conductance of a weak electrolyte at some normality, 
A N , is a measure of the number of ions present in solution, and since A 
is the maximum value, meaning 100% dissociation, the ratio A, V /A is 
the fraction ionized at concentration N. This is also called a, the degree 
ofionization* and the percent of ionization is then lOOa, or 

% ionization = 100(A iV /A ) (5-15) 

Example 5-1. At 1 8 C a 0.001 N soln. of H Ac has a sp. resistance of 25,000 
ohm cm. Calculate the sp. cond. and cc vol. needed to contain a g equiv. wt 
of solute at the given normality; also, the equiv. cond., the equiv. cond. at inf, 
diln., the degree of ioniz., and the percent of ioniz. 

The sp. cond. is the reciprocal of the sp. resistance: 

K = \lp = 1/25,000 = 4.00 x 10- 5 ohm" 1 cm- 1 (Ans.) 

The vol., K, needed to contain a g equiv. wt is 1, 000,000 cc (Ans.) because 
the 10~ 3 N solution only contains 10~ 3 g equiv./liter. Since K is the cond. of 
1 cc, the cond. A N of the g equiv. wt is, 

KV= A N = (4.00 x 10 5 )(10 fi ) = 40.0 (Ans.) 

The value for the equiv. cond. representing 100% dissoc., A , is obtained by 
Kohlrausch's law and the 18 C values of limiting ionic mobilities from Table 

A = AJ + V F= 315 + 35 = 350 (Ans.) 

The ioniz. degree, a, is 

a = A X /A = 40.0/350 = 0.1 14 (Ans.) 

and the percent dissoc. is 

100oc= (100X0.114) = 11.4% (Ans.) 

Molal Conductance and The Identity of Salts 

The molal conductance of a solution is the conductivity of a mole of 
solute at the given molal dilution. If the conductivity of 1 cc of solution 
is measured, one multiplies by the number of cubic centimeters needed 
at that dilution to contain 1 g formula weight of solute. The following 
approximate room temperature data have been found from measurements 
on several 10~ 3 m salt solutions. 

These molal conductance generalizations were of great assistance to 
Werner in the determination of coordination compound formulas. For 
instance, at the above dilution a solution of PtCl 2 -4NH 3 showed a molal 
conductance of 240 and gave 2 equivalents of AgCl when treated with 
excess Ag+. From this he could deduce that the compound contained 



(10~ 3 //j SOLUTIONS) 

Salt Type 

Number of Ions 

Approximate Molal Conductance 



95-1 10 

A 2 B 



A 3 B 



A 4 B 



A 6 B 





three ions, two of which were Cl~, so the complex ion must have been 
[Pt(NH 3 )J*]. 

Calculation of Dissociation by the Freezing Point Method 

It was known to Arrhenius from the previous work of C. Blagden and 
contemporary accurate measurements of F. M. Raoult and E. Beckmann 
that a gram mole of any nonvolatile, nonionized solute dissolved in 
1000 g of solvent (1 molal solution) raises the solvent's boiling point and 
lowers its freezing point characteristic amounts, and that the changes 
are due to the number of particles in solution and not their kind. These 
values for 1 m solutions are called cryoscopic constants. A7" b is the molal 
boiling point constant and AT} the molal freezing point constant, and 
they vary with solvent used. Most of Arrhenius' work was done using 
AT}, by which he showed, for example, that HC1 gave about twice the 
expected lowering a nonionic material should give, Na 2 SO 4 three times 
as much, etc. Such experimental results were excellent support for his 
ionization theory. 



Normal B.P., 

An, c 

Normal P.P., 

AT), C 

Acetic acid 










Carbon disulfide 
















From this type of data one can calculate whether or not a substance 
is a strong or weak electrolyte, and, if the former, then how many ions it 
dissociates into. One can also find the degree of apparent ionization of 
strong electrolytes and apparent molecular weights of any solutes. Weak 
electrolyte ionization cannot be studied easily because it furnishes too 
few ions to effect much freezing point change. 

Example 5-2. A 0.112 molal soln. of BaCl 2 begins depositing ice at 
-0.381 C. Find the apparent % dissoc. 

We may let x be the molality of Ba+ 2 in soln. Then 2x will be the molality 
of CT~, and 0.1 12 x will be the molality of the 'undissoc.' BaCl 2 : 

BaCl 2 = Ba'-' 2 + 2C1~ 

0.112-* . f 2.r 

The total molality of solute particles is (0.1 12 - .r) -f x + 2x = (0.1 12 + 2x). 
When this is multiplied by the f.p. constant, AT,, of the solvent, it should give 
the f.p. drop of the soln., or (0.112 4- 2.r)1.86C = 0.381 C. Solving for x 
gives 0.0465 molal Ba+ 2 . The fraction "dissoc." is 0.465/0.112 and the % is 
100 times that, or 41.5% apparent dissoc. (Ans.) 
How could one check this answer? Verify it by your method. 


1. X-ray diffraction measurements demonstrate that perchloric acid mono- 
hydrate (m.p. 50 C) is a crystalline substance isomorphous (see problem 5, 
Chapter 13) with ammonium perchlorate. This has been offered as evidence 
of the existence of the hydronium ion. Explain. 

2. Tn the chapter, a number of acid-base theories were described and named 
after their discoverers. Five of the more important ones are also known as 
(a) the proton theory (b) the positive-negative theory (r) the electron theory 
(d) the solvent system theory, and (e) the water theory. Match them with the 
men's names and briefly explain why the descriptions are appropriate. 

3. Explain the following reaction in terms of the Lewis theory. Why doesn't 
the Bronsted theory cover it ? 

(CH 3 ) 3 N + BC1 3 = (CH 3 ) 3 NBC1 3 

(The final product is called trimethylamine boron trichloride.) 

4. Select several pairs of values from the first two columns of Table 23-1 
and confirm the data given in the next five columns by calculation. 

5. From Table 5-1, plot for NaCl and K 2 SO 4 , N* versus A v as best you 
can, and extrapolate each line to zero concentration. Check the results from 
Table 5-2 using Kohlrausch's law. Explain the purpose of the graphical 


6. (a) From Table 5-1, calculate the apparent ionization of NaCl, KOH, 
and HC1 at each concentration and present this data in table form. 

(b) Why is the word "apparent"* used in this connection? 

7. (a) According to Table 5-1, at what temperature should O.I and LOW 
NaCl solutions begin to freeze? 

(b) Will the entire solution freeze at the temperatures calculated? If not, 
why not? 

(c) Why do you suppose Arrhenius studied freezing point changes in prefer- 
ence to boiling point changes? 

8. What is peculiar about camphor as a solvent for measuring freezing point 
depressions of its solutions? What is the advantage of using it? 

9. Interpret each of the following by the (a) Lewis theory, (b) Bronsted theory, 
and (c) Usanovich theory: 

(1) Cu+ 2 4- 4NH 3 = Cu(NH 3 )J 2 

(2) HS- + H< = H 2 S 

(3) Ag+ + Cl- = AgCl 

(4) HC1 + A1C1 3 = H+ + 

10. The freezing point of a 1.0 N KC1 solution is -3.35 C. Calculate the 
apparent molecular weight of the salt from this and explain why it differs from 
the actual molecular weight. 

11. Designate the conjugate pairs in the following: 

(a) HN0 2 + H 2 = H 3 0+ + NO 2 

(b) HF + NH 2 OH = NH 3 OH+ + F~ 

12. (a) What is the difference between molal and equivalent conductances? 
(b) A solution of PtCl 4 -2NaCl at a dilution of 1 mole/ 1000 liters has a molal 

conductance of 229. Electrolysis of a more concentrated solution gives no Pt 
or Cl at the negative pole. What can one deduce about the compound's 

13. What would be the approximate freezing point of 0.10 M potassium 
hexacyanoferrate(II)? Explain how this freezing point determination would aid 
in deducing the salt's formula. 

14. (Library) Beckmann invented his thermometer to measure cryoscopic 
changes with an accuracy of 0.002 C. Find reference to this instrument, 
sketch it, and explain how it functions. 

15. One has solutions of the following salts in 0.001 m concentrations. Plot 
a rough graph of molal conductance vs. the charge on the complex ion (scale 
from -f 3 to 3), and explain how the graph would aid one in establishing the 
formulas of other Werner compounds: [Co(NH 3 ) 6 ]Cl 3 , [Co(NH 3 ) 5 Cl]Cl 2 , 
[Co(NH 3 ) 4 Cl 2 ]Cl, [CoCNH^Cy, Na[Co(NH 3 ) 2 Cl 4 ], Na 2 [Co(NH 3 )Cl 5 ], and 
Na 3 [CoCl 6 ]. 

16. (a) A solution made by dissolving 1.28 g of sulfur in 50.0 g of CS 2 begins 
boiling at 46.36 C. Explain. 


(b) The same quantities of solute and solvent are mixed again, but this time 
the solute is phosphorous and the boiling point of the mixture is 46.61 C. 

17. A class determines the f.p. of 1.00 m LiNO 3 to be 3.40 C. J. P. Slipshod 
knows each LiNO 3 can give two ions, so he figures that a f.p. drop of (1.86 C)(2) 
= 3.72 C means 100% ionization. He calculates the apparent ionization in 
the solution by setting up the proportion 100%/3.72C = #%/3.40 C, from 
which x 91.4%. Everybody else in class gets 82.8%, so J. P. figures he is 
the only one who saw through the math. Who is right? 

18. Hydrazine hydrate, H 2 N NH 2 -H 2 O, in aqueous solution turns red 
litmus to blue. Explain. 


1. W. F. Luder, /. Chem. Educ., 22, 301 (1945). (Acid-base theory) 

2. W. F. Luder, /. Chem. Educ., 25, 555 (1948). (Acid-base theory) 

3. C. E. Ronneberg, /. Chem. Educ., 26, 400 (1949). (The Arrhenius theory) 

4. B. Naiman, /. Chem. Educ., 26, 280 (1949). (The Debye-Huckel theory) 

5. T. S. Logan, /. Chem. Educ., 26, 149 (1949). (Acids and bases) 

6. J. T. Stock, /. Chem. Educ., 31, 407 (1945). (Equivalent conductances) 

7. F. P. Chinard, J. Chem. Educ., 32, 377 (1955). (Colligative properties) 

8. R. M. Creamer and D. H. Chambers, Anal. Chem., 26, 1098 (1954). (Conduc- 
tivity device) 

9. W. F. Luder and S. Zuffanti, The Electronic Theory of Acids and Bases, John Wiley 
and Sons, New York (1946) 





Early chemists recognized that not all reactions take place at the same 
speed or proceed to completely consume the reacting materials. It was 
found that some reactions begin rapidly, slow, and finally stop, and it was 
believed that motion of all particles had halted because evidence of further 
action was not visible. It is now known that molecular motion never 
ceases and that the apparent cessation of many chemical processes is due 
to the advent of a reverse combination by reaction products to reproduce 
original reactants. The speed of this reverse reaction is slow at first, 
since not many product molecules are available for reaction, but, as the 
forward reaction increases their number, the probability of their reaction 
increases. When the rates* of the opposing reactions are equal, the 
system comes to its apparent rest. In other words, a state of dynamic 
(instead of static) equilibrium is established between reactants and products 
so no net material transport takes place. The determination of the 
concentration ratio (products/reactants) can be made by quantitative 
analysis of the equilibrium mixture, and under specified conditions is a 
constant for a particular system. As will be seen, this equilibrium constant 
is the starting point for many useful calculations which form the theoretical 
foundations of analytical chemistry. 

Factors Affecting Reaction Rates 

Rates are influenced by a host of factors, only some of which are well 
* Specific rates are expressed in moles per liter reacting per second. 



understood and can be dealt with quantitatively. The following all 
exert effects: the nature of reacting substances, their physical state, 
pressure, the presence or absence of catalysts and their nature, the outside 
energy available such as heat and light, and the concentrations. 

1. Catalytic effects. A catalyst is a substance that affects rates and is 
not changed in the reaction. A catalyst is classed as positive if it accelerates, 
and negative if it slows attainment of equilibrium. Catalysts may function 
in heterogeneous (two or more phases) systems as in the reaction 2H+ + 2e~ 
= // 2 , which is hastened on a surface of finely divided Pt, or in a homo- 
geneous (single phase) system as 2N~ + I~ = 31 ~ + 3N 2 , which is 
accelerated by such catalysts as SCN , S " 2 , or S 2 O^ 2 in solution. Catalysts 
do not alter the final equilibrium state or shift the degree of conversion, 
however. This means that the yield of reaction products is the same with 
or without the catalyst. 

2. Concentration effects. It was discovered in 1 864 by the Swedish 
chemists M. Guldberg and P. Waage that the speeds of many reactions are 
directly proportional to the "active masses" (molar concentrations will 
be used here) of reactants raised to powers which are their coefficients 
in the balanced reaction equation. This statement has become known 
as the law of mass action. For example, to test a solution for acetate, 
one adds a trace of sulfuric acid (catalyst) and some ethyl alcohol. 
Ethyl acetate (identified by odor) and water form, and as their concentra- 
tions increase the reverse reaction becomes important: 

HAc + C 2 H 5 OH = C 2 H 5 Ac + H 2 O 
The specific rate of reaction to the right s r is given by 

* r =[HAc][C 2 H 5 OH]Ar, (6-1) 

and to the left is 

s t = [C 2 H 5 Ac][H 2 0]Ar, (6-2) 

The terms k r and k l are specific reaction rate constants and are different 
for different reactions. They can be determined from equilibrium con- 
centration measurements. The values are constant at constant tempera- 
ture and generally increase as temperature increases. The rate constants 
represent the fraction of molecules that are colliding and reacting per 
second. Therefore, k times a molar concentration gives the number of 
molecules reacting in unit time (1 sec) per unit volume (1 liter). 

The square brackets indicate molar concentrations (see the section 
on concentration conventions, p. 10). If HAc and/or C 2 H 5 OH are in 
abundance, formation of the ester and water will be relatively rapid; 
if instead the latter are in large excess, then acid and alcohol will be 
relatively quickly formed. When s r = s, at some temperature, reactants 


and products are formed in equal amounts each second, so concentrations 
do not change and the system is at equilibrium. 

3. Temperature effects. Reacting molecules must have energy for 
bonding and must approach each other within bonding distance for 
reaction to occur. It is evident that raising the temperature of a system 
will increase both the average molecular energy and the molecular collision 
rate, so in almost every case known, reaction speed increases with increasing 
temperature. For the average reaction at room conditions, it has been 
found that the specific rate increases about 10-30% for every 1 C rise in 

Equilibrium and The Mass Action Law 

Derivation of this law is developed in physical chemistry courses using 
methods from thermodynamics discovered by A. Horstmann (1873), 
J. W. Gibbs (1876), and J. H. van't Hoff(\$86). A much simplified pre- 
sentation will give all that is needed here, however. 

For the general reaction, 

A + B = D + E 

as in the preceding section, the rates right and left are equal to the products 
of molar concentrations times the rate constants: 

At equilibrium, the forward and reverse speeds are equal, 

Collecting like terms, and letting the ratio of constants be K, the equili- 
brium constant, we get 

A- r /A-,= [D][E]/[A][B] = # (6-3) 

By accepted convention, division is always made by putting the resultants 
over the reactants (right side terms over left as the equation is written). 
If the equation contained the coefficients, H', .r, ?/, 2, as 

u'A + a:B = 2/D + zE 
the equilibrium constant's value is 

[B]* (6-4) 

The student may now be surprised to learn that most reactions are 
more complex than the idealized four particle A + B = D + E, because 
a number of intermediate reactions often take place before the final 



products appear. These steps may or may not be known and are not 
deducible without experimental work to find what species may be present 
besides A, B, D, and E, and upon what species the rates depend. Such 
study is the province of chemical kinetics, the science of reaction rates 
and mechanisms, and is too advanced for consideration here. Neverthe- 
less, despite the fact that important omissions were made in the derivation 
of K, the final expression is correct. This is so because the equilibrium 
state is independent of the path by which it was attained, as one can 
deduce from the law of energy conservation. Thus at equilibrium s r 
and s t are equal and equation 6-4 is applicable to all specific equilibrium 
situations to be encountered even though intermediate chemical processes 
might be involved about which we know nothing. 

The important idea is summarized as follows: for a reversible reaction 
at equilibrium at a given temperature, the product of concentrations of 
resultants raised to powers which are respectively their coefficients in the 
balanced reaction equation, divided by the product of concentrations of 
reactants raised to powers which are their respective equation coefficients, 
is a constant. 


FIG. 6-1 . Random distribution of reacting particles showing that if there are .r of one 
type and y of the other, the number of possible reaction collisions is .ry. 

A simple physical picture is this: suppose ten atoms are in a container, 
five of type A and five of B. Assuming equal probability of any A and 
any B colliding and possessing favorable orientation and energy for reaction 
to form AB, the number of possibilities is (5)(5), or, the rate is proportional 
to the product of concentrations, [A][B]. By inserting a rate constant k, 
the actual rate becomes equal to /r[A][B]. 

If the reaction were A + 2B = AB 2 , then, as above, the rate of AB 2 


formation is & r [A][B][B 1]. The last term means that after the first 
particle of B has reacted, there is one fewer B in the mixture, so the 
possibility of further reaction is one less, and the rate will be a little slower. 
If, however, any reasonable quantities of reactants are present, say even 
a millionth of a mole of B, 6.02 x 10 17 particles, the effect of 1 less B is 
insignificant and one writes for the rate, A: r [A][B] 2 . Specific examples 
will later be considered in this way. 

Factors Affecting Equilibrium 

Changes in temperature T, pressure P, and molar concentration A/, can 
cause a shift in the equilibrium points of a system. The principle of 
equilibrium as described by H. L. Le Chatelier (1885) and F. Braun (1886) 
is a qualitative statement of what to expect when such changes are made; 
namely // a change is made in P, T, or M of an equilibrium system, the 
system changes in a direction that tends to compensate for the effect of the 
change. The amount of change cannot be predicted from this statement 
alone, however. 

1. Catalytic effects. Catalysts hasten or retard attainment of the 
state of equilibrium but it is an experimental fact that they do not affect the 
equilibrium point; that is, the ratio of concentrations at equilibrium will 
be the same regardless of how long the reaction takes to get there. The 
positive catalyst thus speeds both forward and reverse rates equally. 

2. Pressure effects. These are mainly important in reactions that 
include gases. For example a solution of carbonic acid decomposes to 
give water and carbon dioxide gas: 

H 2 CO :1 = H 2 + C0 2 

If this system is at equilibrium in a closed container and one increases 
the pressure with a movable piston, the system responds according to 
Le Chatelier's principle in such a way as to absorb the pressure which 
means that the solubility of the gas is increased and reaction goes from 
right to left. Decreasing the pressure (as for example in opening a bottle 
of carbonated water) has the opposite effect. 

3. Temperature effects. If one knows whether heat (expressed in 
kcal/mole) is absorbed or given off in a reaction he can predict by the 
Le Chatelier principle the effect a temperature change will have on an 
equilibrium system. It is known for example that when one dissolves 
silver nitrate in water, the solution absorbs heat and gets cold, or negative 
heat is a product : 

AgNO 3 + H 2 O -> Solution - kcal 
Rewriting gives 

kcal + AgNO 3 + H 2 O -+ Solution 


This shows that if one heats the mixture, the reaction now proceeds from 
left to right; i.e., at the higher temperature the salt has greater solubility. 
That is the direction in which heat is absorbed as the system tries to 
re-establish its previous equilibrium point. At a new temperature the 
process is still a reversible one but the rates of the forward and reverse 
reactions may change different amounts so different equilibrium constants 
are expected at different temperatures. 

4. Concentration effects. Consider the reversible reaction that takes 
place when ammonia is dissolved in water: 

NH 3 + H 2 = NHJ + OH- 

If either the ammonia or water concentrations are increased, the reaction 
velocity toward the right is increased and more ions will appear in solution. 
Adding either ammonium or hydroxide ions shifts the reaction toward the 
left. Thus, addition of a strong base (furnishing many OH~ in solution) 
to an ammonium salt solution will aid one in identifying the latter, since 
NH 3 (which is detected by odor) is liberated. 

Concentration Conventions 

For the purposes of equilibrium discussion, the concentrations oj solids are 
considered to be unchanging. Thev are given the value of unit v and cancel 
from, or simply are not written in, expressions involving the equilibrium 
constant K. For the reaction of a slightly soluble compound dissolving 
in water, as, 

Ag 2 C 2 4 = 2Ag* + QO; 2 (6-5) 

one may write A' = [Ag] 2 [C 2 Oj 2 ]/[Ag 2 C 2 O 4 ] or K[Ag 2 C 2 O 4 ] = [Ag] 2 
[C 2 O 4 2 ], and dropping the solid term, K = [Ag'] 2 [C 2 O- 2 ]. The square 
brackets indicate molar concentrations of the ions. Though this reaction 
takes place in water solution, H 2 O is not a reactant and is not shown. 

For gases, molar concentrations or partial pressures may be used. 
The term K t denotes the equilibrium constant in the former case and K p is 
used in the latter. For example, heating solid CaCO 3 in a previously 
closed and evacuated container gives decomposition, the extent of which 
at equilibrium is expressed by the pressure of CO 2 : 

CaCO 3 = CaO + CO 2 (6-6) 

Since the calcium compounds are solids, the equilibrium statement is 

or K f = [C0 2 ] 

Most systems discussed in the following chapters and illustrated in 
laboratory work concern ions and molecules dissolved in water. Usually 


the solutions are dilute and the use of molar concentrations in equilibrium 
calculations will give results that are close to those obtained from labora- 
tory experimentation. Calculations applying only the simple approach 
thus far described on more concentrated solutions may give quite erroneous 
answers, however, because these solutions do not behave ideally and K 
changes with concentration. An ideal solution in this respect is one in 
which the ions are far enough apart so their random movement is not 
affected by interionic attractions or repulsions. 

For practical purposes this means that only dilute solutions are ideal 
and it was from these that the mass action law was derived. A physical 
picture for one non-ideal system is this: if solid Ag 2 C 2 O 4 is at equilibrium 
with its ions as in equation 6-5, the system behaves ideally because it is 
very dilute. If some NaNO :j is added, ions from the latter will hinder, 
by mechanical interference and ionic attraction, the equilibrium traffic 
between Ag ' and C 2 Oj" 2 on the solid's surface and Ag + and C 2 Oj" 2 dissolved 
in solution, so that more Ag 2 C 2 O 4 must dissolve to re-establish the original 
reaction velocity. 

K actually varies with the valence, size, and population of ions in the 
solution, and in refined calculations* cognizance is taken of these factors, 
though we will ignore them. The equilibrium constants used in problems 
here are all of "classical," rather than thermodynamic, derivation. 

Partition an Example of Heterogeneous Equilibrium 

Heterogeneous equilibria systems are those containing two or more 
phases, several examples of which have been mentioned. One more is 
given here in greater detail because of its importance as a method for 
quantitatively studying the equilibrium phenomenon. 

If two immiscible liquids arc in contact, and if a common solute is 
added and the mixture shaken, the solute distributes itself between the 
two solvents in the ratio of its solubility in each. It is capable of doing 
this because its molecules are moving in all directions and pass through 
the liquid-liquid interface. 

Suppose Br 2 is shaken with a mixture of two immiscible solvents, 
H 2 O and CC1 4 , until no further concentration changes are found. The 
organic layer is noted to be a darker brown than the aqueous layer, 
showing that Br 2 is more soluble in CC1 4 . If s l is the rate at which Br 2 
leaves the water layer, s l is proportional to [Br 2 (H 2 O)], the concentration 
of Br 2 in water, and equal to this times a rate constant, k^: 


* See any physical chemistry text on "activities," "activity coefficients," and "ionic 


In the same way, if s 2 is the rate at which Br 2 leaves CC1 4 , s z is equal to the 
concentration of Br 2 in CC1 4 multiplied by another rate constant: 

J 2 = A: 2 [Br 2 (CCl 4 )] (6-8) 

At equilibrium the two rates are equal and collection of terms gives an 
equilibrium constant which is called the distribution (D) constant or 
partition coefficient:* 

A'iM' 2 = K D = [Br 2 (CCl 4 )]/[Br 2 (H 2 0)] (6-9) 

Experimental determination of the layer concentrations will give the 
value of K u , which may then be used to predict partition when different 
weights and volumes of the three substances are mixed. The value of this 
K t) at 20 has been found to be 23. 

See Example 9-3 for an illustration of how K n can be used in solving 
equilibrium problems. 

The Dissociation of Water -an Example of Homogeneous Equilibrium 

Water ( W) ionizes slightly and this reversible process is one of the most 
important ones in chemistry: 

2H 2 O = H 3 O+ + OH or H 2 O = H+ + OH 

The "concentration" of water is considered constant in dilute solutions 
and is incorporated in the equilibriums constant: 

K = [H'][OH ]/[H 2 0] or K[H 2 O] = [H'][OH-] 

and the water constant has the definition, 

K ir = [H+][OH ] = 1 X 10 14 (6-10) 

This constant can be determined from measurements on salt hydrolysis 
(Chapter 10), electrochemical cells (Chapter 11), and electrical conduc- 
tance. The last method involves the preparation of water of the highest 
purity by means of ion exchange resins and low pressure distillations to 
remove all impurities including dissolved gases. This is followed by 
finding the conductance, as explained in Chapter 5 and in special experi- 
ment 1. The solution is dilute enough to consider it at infinite dilution. 
At 25 C the conductance of a centimeter cube of H 2 O is 5.53 x lO" 8 
ohms" 1 cm" 1 . The equivalent conductances at infinite dilution are 
H+ = 348 and OH- = 192. The cc of H 2 O thus contains 5.53 x 10~ 8 / 
348 + 192 = 1.00 x 10~ 10 g equivalents of H+ and of OH~. A liter 

* The convention here is to put the concentration in the extracting phase in the 
numerator and the concentration in the extracted solution in the denominator. 


of water at 25 C consequently contains H+ and OH~ at 1.00 x 10 7 
molar concentrations, or from equation 6-10, 

[H'][OH-J = K n = (1.00 x 10 7 ) 2 = 1.00 X 10~ 14 (6-11) 

K n increases with an increase in temperature; it is 0.61 X 10~ 14 at 18 C 
and 50 x 10 14 at 100C. The value 1.00 x 10~ 14 will be used in our 

pH, pOH, pK n 

Because small molar concentrations of H 1 and OH" are common, 
S. P. L. Sorenson in 1909 proposed that a convenient way to express them 
would be as their negative logarithms. This is denoted by the small 
letter /?. The relationships (in fairly dilute solutions) from equation 6-10 

pH = - log [H*] = log (1/[H<-])* (6-12) 

[H+] = 10-* 11 (6-13) 

pOH = - log [OH-] = log (1/[OH-]) (6-14) 

[OH-] = 10 ' OH (6-15) 

pH + pOH = p/C ir = 14.00 (6-16) 

P K ]r = -logtf, r (6-17) 

Due to the equilibrium expressed in equation 6-10, it is seen that when 
a solution contains equal numbers of H + and OH", the pH at room 
conditions will be 7.00 and such solutions are called neutral. If [H+] 
> 10~ 7 A/, />H < 7, and the solution is acidic. If [H+] < 10 ' 7 M, then 
there is more OH~ than H 1 since the equilibrium must be maintained, so 
pH > 1 and the solution is basic (alkaline). The usual limits of the pH 
scale are from 1 M H+ (10 U M OH) to 1 M OH- (10~ 14 M H+), or 
from pH (/>OH 14) to pH 14 (pOH 0). 

pOH 14 








[H+l 1 10"' 10 ' 10 "M 

| OH"] 10" 14 10" 7 10" 5 1M 

FIG. 6-2. The/?H scale showing a few representative values. 

Example 6-1. One dissolves 0.0092 g of Ca(OH) 2 (mol. wt 74) to make a 
liter of soln. Calculate [OH~], [H+}, /?H, and pOH. 

* See Appendix A3 on handling logarithms. 


The molarity of the soln. is 9.2 x 10~ 3 /7.4 x 10 = 1.24 x 10~ 4 . One may 
assume this dilute, strong base is completely dissoc. so 

[OH-] = (2)(1.24 x 10~ 4 ) = 2.48 x 10~ 4 M (Ans.) 

From 6-10, 

[H+][OH-] = 1.00 X 10~ 14 , [H+] = 4.03 x 10~ u M (Ans.) 

Then from equation 6-12 

pH = - log (4.03 X 10- 11 ) = - 0.60 + 1 1.00 = 10.40 pH (Ans.) 
pOH is found from either equations 6-14 or 6-16 to be 3.60 pOR (Ans.) 

Example 6-2. Methyl orange added to an HC1 soln. gives a yellow-orange 
color. Approx. the formality of the HC1. 

From the appendix table of indicators, A17, pH ^ 4.4, so 4.4 = log [H+], 
or [H+] = 10~ 4 - 4 . Since log tables do not provide for negative mantissas, the 
number is written as [H+] = 10~ 5 x 10- 6 = 10- r> x antilogO.6 = 4 x 10 5 for 
both the hydrogen ion molarity and hydrochloric acid formality. (Ans.) 

pH Measurement 

pH is determined most often by indicators or a pH meter, as illustrated 
in special experiment 2. 

The water equilibrium and the effect of pH on several systems will be 
developed in the next chapters. % 


1 . Write expressions for the equilibrium constant K for each of the following, 
utilizing only those substances whose concentrations may change: 

(a) N 2 + 3// 2 = 2A7/3 + 34 kcal 

(b) 4//C/ + 2 = 2H 2 O + 2C/ 2 
(r ) C + 2 = C0 2 + 94.4 kcal 
(cl) 7/2 + 7 2 = 2H1 + 12 kcal 
(e) BaSO 4 = Ba+ 2 + SO 4 2 


(/) HSO^ = H+ + S0 4 

(g) H 2 + C 2 4 - 2 = HC 2 4 - 4- OH 

(h) HN0 2 = H+ -f N0 2 - 

(i) CuBri" 2 = Cu+ 2 -f 4Br- 

(j) NH^ + H 2 = NH 3 + H 3 0+ 


2. Kp for CaC! 2 -2H 2 = CaCI 2 + 2H 2 O is 10~ 8 atm at C. Explain. 

3. For reactions (a), (c) and (d), in problem 1, explain the effect of making these 
independent changes: (a) add heat (b) lower the total pressure (c) remove the 
products as formed (d) add one of the reactants, and (e) pump an inert gas like 
helium into the otherwise closed reaction chamber. 

4. The all-gas system in (d) of problem 1 was one which was carefully studied 
when equilibrium concepts were originally being formulated (Bodenstein, \ 894). 
A sealed reaction flask mixture of 0.9151 g of I 2 and 0.0197 g of H 2 was allowed 
to come to equilibrium in a bath of boiling sulfur (444.6 C), then quenched in ice 
to retain the equilibrium concentrations. The vessel was broken open, the 
HI and I 2 absorbed in aqueous solutions, and the H 2 collected in a gas buret. 
Corrected to standard conditions, the H 2 had a volume of 143 cc. Calculate 
the moles of each constituent at equilibrium and K c . 

Why doesn't flask volume matter in this calculation? 

H 2 = 0.00638, HI = 0.00694, 1 2 = 0.00014, K c = 53.5 (Ans.) 

5. Using K f from problem 4, calculate the equilibrium concentrations in a 
2-liter-flask at 444.6 C into which originally was placed 10 g of I 2 and 0.059 g 
ofH 2 . 

6. J. P. Slipshod calculates the /?H of 10~ 8 F HC1 as follows: 

10~ 8 FHC1 = 10 8 A/H+ pH = - log 10~ 8 = log 10 8 = 8/?H 

He then remembers that a pH greater than seven means a basic solution and he 
cannot figure out why HC1 should be basic. What happened? 

7. A mixture of 10.0 g of HAc and 8.00 g of C 2 H 5 OH is held at 45 C until 
equilibrium is established. Analysis shows 2.00 g of H 2 O and 9.68 g of 
CH :i CO 2 C 2 H 5 are present (see reaction above equation 6-1). (a) Find K. 
(b) If a mole of each reactant had been present initially, what weights of products 
would have been produced? 

8. Organic chemists are interested in the partition application of chemical 
equilibrium since if they wish to extract a compound from one solvent with 
another immiscible solvent, they are able to calculate the efficiency of the process. 
If A/o = original weight of a solute in water, M n = final weight of solute in 
water after n extractions, K D = the distribution constant, V o the volume of 
organic solvent used in each extraction, and V w = the volume of water solution 
it can be shown that 

M n = MJi(V w lK D )l(Y + V w lK D )} n 

It is known that Br 2 is 80 times more soluble in CS 2 than in H 2 O and that these 
solvents are practically immiscible. Calculate: (a) K D . (b) The mg of Br 2 
left in the water layer if 200 cc of H 2 O containing 40 mg of Br 2 are shaken with 
30 cc of CS 2 . (c) The mg of Br 2 left in the water layer under the same condi- 
tions, except this time three extractions with CS 2 are made, using lOcc each 
time, (d) What generalization can you form by comparing the results from 


9. Complete the table for solutions A, B, C, and D. 














3.2 x 10~ 5 

10. (a) A solution has a /?H of 1.0. Explain, (b) Could any solution 
have a pR > 14? Explain, (c) What does 10~ pH equal? 

11. J. P. Slipshod reads that K w is 50 x 10" 14 at 100 C and calculates the 
pH of boiling water to be 6.15. This gets both him and his Uncle Frisbee 
excited enough to form a corporation for the purpose of cleaning scale from 
boilers without the use of expensive chemicals or manual labor. J.P.S. and 
U.F. reason that if hot water is run through the tubes it should dissolve car- 
bonates because its/?H indicates it is acidic. On their first job they circulate hot 
tap water for six days without visible results and are shown the door. J.P. 
then alibis that salts in the tap water threw off his computations and distilled 
water will be used on their next caper. What about this ? 

12. Consider the reaction in equation 6-6 to be at equilibrium. What is the 
effect of adding more CaO? Explain. 


1. J. H. Hildebrand, /. Chem. Educ., 23, 589 (1946). (Introduction) 

2. N. Hackerman, J. Chem. Educ., 23, 45 (1946). (Introduction) 

3. A. B. Martin and J. C. Speakman, ibid., 27, 272 (1950). (Activity coefficient) 

4. G. S. Forbes, J. Chem. Educ., 27, 402 (1950). (Inorganic mechanisms) 

5. D. D. Deford, J. Chem. Educ., 31, 460 (1954). (Reliability of equilibrium 

6. D. MacRae, /. Chem. Educ., 32, 172 (1955). (Thermodynamics) 



In the last chapter it was stated that chemical equilibrium principles 
can be broadly applied to soluble weak electrolytes in solution but not to 
strong ones. The former ionize gradually in a predictably progressive 
fashion with dilution, but the latter, already largely in the form of freely 
moving and individually functioning ions at moderate concentrations, do 
not exhibit the same regularity. This is easily demonstrated. 

A Weak and a Strong Base Compared 

By the method described in Chapter 5 and illustrated in special experi- 
ment 1 , one may obtain conductivity data on various dilutions of potassium 
and ammonium hydroxides, for example. From this one can calculate 
the equivalent conductance, A, for each compound at each dilution, and 
get the figures shown in the accompanying table. By knowing that at 




A, 18 C 

NH 4 OH 

A, 18 C 













infinite dilution the limiting ionic conductance of NH+ = 64, K^ = 64, 
and OH = 174 and applying Kohlrausch's law as in Chapter 5, one 
finds that both compounds have the same equivalent conductance at 
infinite dilution, A = 238. The ratio A/A is the apparent degree of 
dissociation or fraction of substance in the free ionic form. This is shown 
in Table 7-2. Both KOH and NH 4 OH apparently increase in their 




A/A O 

NH 4 OH 











amounts of ionization but the two are not similar phenomena, as will be 

The loni/atiou ,/j NH 3 in H 2 O 

Ammonia, NH 3 , dissolved in water forms hydrates, but there is no way 
at present to differentiate between NH 3 -H 2 O and NH 4 OH. While the 
latter has not been proven to exist, it is convenient to use it in equation 
writing and the student should understand that these two expressions are 

NH 3 + H 2 O = NH+ + OH- 

NH 4 OH = NHJ + OH 

The latter will be used most often here. 

Considering the 1.0 M NH 4 OH solution first, from Table 7-2 one sees 
that the degree of dissociation is 0.00374. This is the fraction of a mole 
which is ionized, so the concentration of NH^ is (0.00374)(1 mole) 
= 0.00374 M. Since OH is formed in equal amount, its concentration 
is also (0.00374)(1 mole) = 0.00374 A/. The NH 4 OH left undissociated 
is the difference between unity and that dissociated, 1.00 0.00374 or 
0.996, and its concentration is (0.996X1 mole) = 0.996 M. In the general 
case when a, the degree of dissociation, is known one finds the concentra- 
tions of the various species in solution by multiplying a and 1 a by 
the appropriate molarity. It is handy to put these values beneath the 
respective terms of the reaction equation for reference in problem solving. 
Thus for some weak electrolyte HQ. which gives ions H ' and Q~ in solution 


and whose degree of ionization is a at an HQ concentration of A/, one 

HQ % H+- + Q- (7-1) 

(1 a).V M-x. Mi 

For the particular case under study, 

NH 4 OH ^ NH 4 f + OH- (7-2) 

(i) (ooo374)0) 

The ionization constant, K n , for the base may be found from these figures, 
and as was pointed out is a special case of the general equilibrium constant. 
The concentration of water is essentially constant since a large excess 
exists in dilute aqueous solutions, and it will be considered incorporated 
in the equilibrium constants but not appear as such: 

K n = [NH 4 <][OH-]/[NH 4 OH] = (0.00374) 2 /0.996 = 1.40 x 10 5 (Ans.) 

At 0.10 A/, the degree of dissociation (fraction ionized or dissociated) 
is 0.0139 and the undissociated fraction is 1 minus this value or 0.986. 
As above, one obtains the equilibrium concentrations of all species, then 
solves for K n : 

NH 4 OH ^ NHJ + OH 

(09Stt)(0.1) (0.0130)(0 

K B = [(0.01 39)(0. 1)] 2 /(0.986)(0.1) = 1.96 x 10~ 5 (Ans.) 

At 0.01 M the same reasoning leads to 

K n = [(0.0403)(0.01)] 2 /(0.959)(0.01) = 1.70 X 10 5 (Ans.) 

The three results for K h , are in fair agreement with one another and in 
other dilute solutions at the same temperature, the result of more measure- 
ments gives K n values clustering about 1.8 x 10 5 . One concludes that 
the chemical equilibrium law is suited to describe the behavior of this 
typical weak electrolyte. (These calculations are all approximate: three 
significant figures and many times only two are justified.) 

The picture one should have of this type weak electrolyte in solution is, 
molecules ^ ions, the equilibrium shifting toward the right with dilution. 
One deduces this from Le Chatelier's principle: dilution decreases the 
ionic concentration so the molecules dissociate further to make up the 
deficiency and restore equilibrium. It is only at high dilution that this 
approaches 100% or that a _^: 1. 

The Ionization of KOH 

If one assumes that KOH behaves as does NH 4 OH and that con- 
siderable undissociated KOH is present in solution along with its ions, 
KOH ^ K+ + OH~, calculations can be made in the same way. At 


1.00 A/, K H = 2.65, at 0.10 M, K lf = 0.764, and at 0.01 M, K B = 0.218. 
These results show a decrease of K Jf with dilution and one concludes that 
chemical equilibrium principles do not apply in this case. The failure is 
due to the fact that KOH as a typical strong electrolyte, is ionic all the 
time but in fairly concentrated solutions the ions are too close together to 
completely escape each other's attraction and so the conductivity is lower 
than expected. We thus apply equilibrium calculations only to weak 
acids and bases in this chapter. 

Weak Monoprotic Acids 

The mathematical treatment of weak acids of the type HQ is similar 
to that shown for NH 4 OH and the general results and conclusions are 
the same. Again increased ionization compensates for increased dilution 
over a wide concentration range and K t , the ionization constant for a 
weak acid, is a constant. 

K A for acetic acid, (CH 3 CO 2 )H (abbreviated hereafter H Ac) is 1 .8 x 10~ 5 
at 25 C, showing it to be comparable in strength (or weakness) as an acid, 
to NH 4 OH as a base, and solutions of the same concentration of these 
substances contain equal numbers of ions. Suppose one wishes to know 
the per cent of ionization of 0.2 M HAc solution. The problem can be 
solved in several ways. 

Example 71. (Method 1 ). If a is the degree of dissoc. of HAc into ions, 
then (1 a) will be the degree of nondisSoc. The amount of H+ and Ac~ at 
equil. will each be 0.2oc and the concn. of HAc will be (0.2)( 1 a) as in equation 

HAc <=* H + Ac (7-3) 

(0.2)(1 -a) 0.2* 2a 

K A = [H+][Ac-]/[HAc] = 1.8 x 10~ 5 = (0.2a) 2 /(0.2)(l - a) (7-4) 

Collecting terms and dividing through by 0.4, one obtains a 2 + 9.0 x 10~ 5 oc 
9.0 x 10 5 = 0. This is a quadratic equation of the form, 

a 2 + b(x. - b = (7-5) 

which is common in these problems but for which there is generally no simpler 
solution than the quadratic formula 

a = - b -t (b - 4ac)*/2a (7-6) 

where a, b, and c are coeff. of the general quadratic ay, 2 + ben + c = 0. 

Usually in our calcns. a = 1 and b = r, as above. 


a = - 9.0 x 10- 5 [(9.0 x 10~ 6 ) 2 - (4) (1)(- 9.0 x 10- 5 )]*/(2)(l) 
a = - 9.0 x 10~ 5 - 1.9 x 10~ 2 /2^ - 0.95 x 10~ 2 
and a = - 9.0 x 10~ 5 -f 1.9 x lQ- 2 /2 ^ + 0.95 x 10' 2 


The latter root is the only one with significance since a neg. degree of ioniz. is 
impossible. The lowest value a can have is 0, the highest , 1.00 (why?). This 
degree of ioniz. is convertible to % ioniz. since a is the decimal equiv. of the 
per cent, or lOOoc = %. The ioniz. in this soln. is 0.95%. (Ans.) 

Example 7-2. (Method 2). Let x = the molar concns. of both H + and 
Ac" since these are formed in equal quantity. Then (0.2 x) == the molar 
concn. of HAc left undissoc. 

HAc *=; H+ + Ac- (7-7) 

(0.2 -j) x x 

K A = 1.8 X 10~ 5 = x 2 /(Q.2 - ,r) (7-8) 

This leads to another quadratic from which one obtains 
x = 1.9 x 10~ 3 M H+ and Ac~ 

The fract. ionized is 1.9 x 10 3 /0.2 = 9.5 x 10 3 (this is a from the first 
method), and one again finds the dissoc. to be 0.95 %. (Ans.) 

Example 7-3. (Method 3.) Either of the above methods may be simplified 
by the assumption that weak electrolytes are not going to ionize greatly at 
concns. usually employed. So one may say in Method 1 that if a is small, 
(1 a) ^ 1 ; and in Method 2 if # is small, (0.2 .r) ^ 0.2. This assumption 
is often made when the % ioniz. is going to be < 10% or the concn. is > 10~ 3 M. 
Going back to Method 1, the new values are, 

HAc % H+ 4- Ac ' (7-9) 

(O.'2)(l) 0.2a 0.2a 

K A = 1.8 x 10- 5 ^(0.2a) 2 /0.2^0.2a 2 
a ^ (1.8 x 10- 5 /0.2)l ^ 9.5 x 10' 3 as before. (Ans.) 

Since the same approx. is made each time one works this type problem, the last 
step can be generalized, 

a = (K ,/C)* (7-10) 

where C is the original molar concn. of weak acid or base. 
Applying the approx. to Method 2, gives 

HAc ^ H+ + Ac~ 

K A ^ 1.8 X 

3D* [(0.2X1.8 x I0~ 5 )]* = 1.9 x 10- 3 as before. (Ans.) 
A generalization* can be discovered in the last step again by recognizing that 

* Formulas of general interest for solving type problems will be developed to call 
attention to reasoning processes and to show relationships that might be overlooked if only 
isolated, specific examples are given. The student is not encouraged \ however, to lean 
heavily on memorized equations or set patterns for other than simple problems. It is 
always better to spend time learning the reasoning and principles involved, since these are 
the best tools for attacking the "different" problem, where no obvious solution is at once 
evident. It is the problem where no formula is given that separates thinkers from memorizers. 


0.2 is the weak acid's original molar concn. and 1.8 x 10~ 5 is the ion constant. 
Using the abbreviations as above, 

[H+]-=(C.K )* (7-12) 

The approximate solutions are as good in these examples as the more 
exact ones since both jc and a are small. At higher dilutions more 
difference will be noted, but for most purposes this shorter method is 

pH of Weak Acid and Base Solutions 

In the last two sections, enough data was presented in the examples to 
have calculated the pH of the NH 4 OH and HAc solutions, since the 
ionization constants and the concentrations were known. This is worthy 
of illustration. 

Example 7-4. Calculate the [OH }, [H+], pU and pOH of 0.30 M NH 4 OH. 

If we let x = [NH 4 ] = [OH~], and if we assume the concn. of those ions is 
insignificant compared to 0.30 M, then the concn. of NH 4 OH is not altered by 
the dissoc. 

K n = 1.8 x 10~ 5 = fr-r/OJO, and x = 2.32 x 1Q- 3 M = [OH~] (Ans.) 

From the water constant (Chapter 6) K w = [H+][OH-] = 10 x 10~ 15 , [H+] 
then is found to be 4.32 x 10~ 12 M. pQH and/;H are respectively - log [OH ] 
= 2.63 and log [H+] = 11.37 (Ans.) 

The latter may also be found from the identity, pH + pOH = 14.00. 

If one is working with an unknown weak acid or base whose concentra- 
tion and pH are known, K A or K R can be found. This is of interest to 
research workers who have need of characterizing new substances. (See 
special experiment 2.) 

Example 7-5. A new, weak monoprotic acid HQ, has been discovered. 
When 2.304 g are dissolved to make 50.00 ml of solution, the pH of the mix. as 
measured with a/?H meter is 2.45 and 35.40 ml of 0.2640 N NaOH are equiv. to 
this amount of acid in a titration. Show that the easily obtained data leads to 
calcn. of the mol. wt of HQ, K A , and a. 

From V A N A = V n N n (Chapter 2) one finds the normality of the HQ soln. 
N x = (35.40)(0.2640)/50.00 = 0. 1 869 N. Since the acid is monoprotic, N = M. 
The acid soln. contained 2.304 g/50 ml, which is equiv. to 46.08 g/liter. This 
means that 46.08 g is 0. 1 869 part of a g mole or the mol. wt of HQ is 247. (Ans.) 

From the/?H of 2.45, one calcs. the [H f ] to be 3.55 x 10~ 3 M, and since one 
Q- is formed with each H + , [Q-J = 3.55 x 10" 3 M also. 

The equil. concn. of HQ is the original concn. minus the quantity dissoc. 
or 0.1 869 - 0.00355 = 0.1834 M. The fraction dissoc., a, will be 0.00355/0. 1869 
= 1.90 x 10" 2 (Ans.) and the ioniz. constant will be, 

X A = [H+][Q-]/[HQ] = (3.55 X 10- 3 ) 2 /I.83 x 10* 1 = 6.89 x 10 5 (Ans.) 


Some Electrolytes Compared 

The magnitude of K values for weak electrolytes and the pH of their 
solutions depends upon the extent of dissociation as just illustrated. The 
dividing line between strong and weak electrolytes is somewhat arbitrary, 
but those which obey the mass action law when dilution is the only 
variable are classed as weak, and most of the common ones are listed in 
Tables A20 and A21 with their K icm values.* In general they have ion 
constants smaller than 10 ~ 2 and their 0.1 M solutions are less than about 
30% ionized. From those tables and the methods given in this chapter, 
one can calculate such things as the % ionization and/?H of weak electrolyte 
solutions and find the changes with changes in concentration and AW 
For example, K A for HSO^ is 1.26 x 10~ 2 , />H of a 0.1 M solution is 
found to be 1.53, and the acid is 29.6 % ionized, whereas K A for HCN is 
4.0 x 10~ 10 , pR of a 0.1 M solution is 5.2, and it is 0.0063% ionized, etc. 
The student should verify these calculations, given concentration and K A . 

Some generalizations concerning the strength of oxygen acids can be 
made from the A^n table and are useful to know as a means of getting 
first approximations. Acids of the type H, f XO,, (such as H :t BO 3 ) in which 
a is an integer and X generally a nonmctal, usually have first ion constants 
less than 10~ 7 . Acids of the type H,,XO, m , as HNO 2 , are moderately 
weak and have first ion constants of about 10~ 3 . Strong oxygen acids 
having first ion constants approximately equal to 10 3 have formulas 
H,,XO, lf2 , as HNO :l , and there are a few very strong acids, A: ion 10", 
such as HC1O 4 , whose general formula is H,,XO,, +3 . 

Effect of Foreign Ions on Weak Acid and Weak Base Equilibria 

So long as the weak electrolyte is alone with its ions in aqueous solution, 
the type calculations illustrated thus far handle the possible problems. 
If other substances are present, however, they may affect the dissociation. 
Three cases arc encountered. 

1. Indifferent. Nonreacting Ions. If KC1 is added to an HAc solu- 
tion, the ions K 1 and Cl are "indifferent" or "uncommon" to the original 
system. Offhand it would seem that they should have no effect on HAc 
= H* + Ac~, but, as pointed out in the preceding chapter, any added 
ions interfere with the weak electrolyte's equilibrium and decrease the 
effective H + and Ac~ concentrations by interionic attractions that hinder 
the latter's recombination. Therefore HAc must dissociate to a greater 
than normal degree to refurnish the solution with equilibrium ion concen- 
trations. /Cion will thus be larger than normal. This is known as the 

* tfii, n is the general term for the ionization constant. K A will be used to denote the 
ionization constant of a weak acid and K n used for weak bases. 


indifferent or uncommon ion effect. Enough is known of this effect so 
that in advanced work correction can be made for it, but we will not 
pursue it further. 

2. Indifferent, Reacting Ions. If to a weak base is added an acid, 
or to a solution of weak acid is added a base, the added substance contains 
ions indifferent or uncommon to the original solute but capable of reacting 
with it. If the added substance gives a precipitate or forms a complex 
with the original, a reaction again occurs that will alter concentrations. 
Some examples of this are, 

HCN + OH- = H 2 + CN- 

HCN + Cu+ = CuCN + H+ 

4HCN + Ni+ 2 = Ni(CN)j- 2 

In all of these, the original HCN concentration is decreased and its ioniza- 
tion promoted because a demand is created by the added reagent for H f 
or CN~. Some reactions of this type will be included in further discussion. 
3. Common Ion Effect. One can predict from Le Chatelier's prin- 
ciple, that if ions common to those found in a solution of weak acid or 
base are added to it, ionization of the weak electrolyte is inhibited. Thus 
adding either HC1 or NaCN to a solution of HCN increases respectively 
the concentration of H < and CN , shifting the reaction HCN '=? H* + CN - 
to the left. The reason for this is that the system was originally in 
equilibrium with a much smaller quantity of its ions (K A = 4 x 10 10 ) 
and now must readjust to the new conditions. If H ! is added, CN - will 
react with it. This increases the HCN concentration proportionately, 
and when equilibrium is again attained, the [CN~] will be very small. 
If CN- is added, H 1 will react with it to increase the HCN concentration 
again, and incidental to this will be a rise in the pH caused by a dis- 
appearance of H + . These changes are due to the common ion effect 
which is the most important effect of the three mentioned. Application 
is made in buffer solutions. 

Simple Buffer Solutions and the Common Ion Effect 

A solution of weak acid(s) or weak base(s) plus salt(s) oj weak acids or 
bases is called a buffer solution. Frequently the salt is a derivative of the 
same acid or base in the solution. By furnishing a comparatively large 
quantity of common ion to the solution it depresses the [H 1 ] from the 
weak acid or the [OH~] from the weak base by the common ion effect. 
These solutions come to equilibrium at some definite p\\ and despite 
small additions of strong acid or base are able to maintain that pH 
through equilibrium shifts. Such solutions are said to be buffered against 


pH change, though large additions of strong acid or base will use up the 
buffer capacity. Buffer systems are important in biochemical processes 
(blood contains the system H 2 CO 3 HCOj ) as well as in certain laboratory 
procedures where pH regulation is needed during chemical reactions. 

Buffers may be classed as acidic buffers, weak acid plus a salt, and as 
basic buffers, weak base plus a salt. A typical basic buffer is the system 

NH 4 OH = NH+ + OH 
NH 4 C1 = NH+ + 


ci- ) 

and a typical acidic buffer is 

HAc = H+ + Ac- } (7-13) 

NaAc = Na+ + Ac-j (7-14) 

Consider adding a little HC1 to a solution containing the last pair. 
The increase in H f disturbs the system (7-13) since [H+] is momentarily 
abnormally high. Equilibrium is soon reestablished however as Ac- 
combines with H+ to produce more HAc. There is plenty of Ac~ available 
in solution for this purpose from the reaction shown in equation 7-14. 
In this way, IV is used up and the buffer's pH is not much affected. 
Consider adding base to the same system. The OH~ is neutralized by H+ 
(see equation 7-13), causing the reaction to shift right and give more H^ 
since the original equilibrium H+ concentration was greater. Such shift 
is possible because acetic acid is present primarily as undissociated mole- 
cules and much H+ is stored in that form, ready for release if the solution 
pH is raised. Again, the added substance is nullified as a pH disturbance. 
The student should consider similar reactions with the basic buffer. 

For our purposes, approximate solving of buffer problems will be 
satisfactory, and the following approximations are useful in simplifying 
the calculations. One assumes that 

(1) Since the solutions are dilute, molarities can be used validly and 
K A remains constant. The uncommon ion effect will not be considered. 

(2) The salt in the buffer is 100% dissociated into free ions. Since the 
weak acid or base is negligibly dissociated, the common ion concentration 
will therefore be calculated from the salt concentration. 

(3) If a small amount of strong acid or base is added to the buffer 
mixture, the strong acid or base is completely used up, leaving a buffer 
solution similar to the original but of different concentration. Since the ' 
solutions are dilute, volumes are additive. 

Example 7-6. A soln. is 0.10 M in HAc and NaAc. To 90 ml of this is 
added 10 ml of 0.01 M HC1. Calculate (a) pH of original buffer and (b) pH of 
final mixt. 


Assuming that the weak acid is so little ioniz. that the concn. of HAc molecules 
in 0.1 M HAc is about the same as in the buffer soln., namely [HAc] ^0.1, 
and that the common ion, acetate, is furnished almost entirely from the salt 
which exists as free ions, then [Ac~] ^0.1. Knowing K A for HAc, only [H+] 
is unknown so 

1.8 x 10- 5 = [H+][Ac-]/[HAc] = [H+](0.1)/(0.1) 

[H+l = 1.8 x 10- 5 M, from which pH = 4.75 (Ans.) 

Mixing this buffer with HC1 has two effects; first, the solns. mutually dil. 
each other and second, the addn. of H+ momentarily unbalances the equil. so 
that H+ and Ac~ react to form HAc and reestablish equil. concns. 

At equil., the mixt. vol. is 100 ml and the concn. of HAc will reflect both the 
fact that 90 ml of 0.10 M HAc is dil. to 100 ml and that 10 ml of 0.01 M HC1 
dil. to 100 ml produces some HAc, by H+ + Ac~ = HAc. Its equil. concn. 
is [HAc] = (90/100X0.1) + (10/100)(0.01) = 0.091 M. The equil. acetate concn. 
will be approx. the original minus the amount used by H+, or reasoning as above 
[Ac~] = (90/100X0.1) - (10/100X0.01) = 0.089 A/. Note the fundamental 
assumption is that H+ + Ac~ = HAc is a 100% conversion. Again the 
hydrogen ion concn. is the unknown in the equil. constant expression for the 
weak acid : 

1.8 x 10- 5 = [H+](0.089)/(0.091) 

from which pH is found to be = 4.73 (Ans.). 

Thus despite addition of strong acid the buffer pH is essentially un- 
changed. The same type calculation shows buffer mixtures equally well 
able to compensate for additions of strong base. Had the added strong 
acid or base been 0.1 M instead of 0.01 M in this problem, one can show 
that [H+] will still only change about 20%. Three related points worthy 
of consideration are given in the next sections. 

pH Changes in Unbuffered Solutions 

In solutions that contain no substances that can react to use up H+ or 
OH~, pH changes are great upon strong acid or base addition. 

Example 7-7. To 90 ml of H 2 O are added 10 ml of 0.01 M HC1. Calculate 
the initial and final pH. 

The pH of water is 7.00. (Ans.) After mixing, the [H+] in 100 ml is 
(10)(0.01)/100 = 10~ 3 Af, and the/>H by inspection is 3.00. (Ans.) 

Compare this change of 4/?H units with the change of 0.02 units from 
Example 7-6. (Why is there such a difference?) 

Buffer Range 

The effectiveness with which a buffer system resists pH change is directly 
proportional to its concentration. In general laboratory practice, these 


solutions are employed in the range 0.02-2.0 M with a ratio of weak 
electrolyte to salt in the range 1/10-10/1. While moderate dilution will 
not alter the solution /?H, the ratio of weak electrolyte to salt will. 

Example 7-8. Calculate the useful pH range covered by the buffer system 
salicylic acid, H(C 7 H 5 O 3 ), a weak monoprotic acid, (K A = 1.1 x 10~ 3 ) and its 
sodium salt, Na(C 7 H 6 O 3 ). 

Let us abbreviate the radical Sa~, then 

K A = [H+][Sa"]/[HSa] or [H+] = K A [HSa]/[Sa-] 

Since almost all Sa~ comes from the salt, if we let 
[A] =s acid concn. and [S] = salt concn. then in a general case, 

[H+] = ([A]/[S])K 4 (7-15) 


pH c* - log [([A]/[S])KJ (7-16) 

If the ratio [A]/[S] is 1/10, 

/?H = - log (1/10X1.1 x 10~ 3 ) = 3.96 (Ans.) 

and if that ratio is 10/1, then 

pH = - log (10/1X1.1 x 10~ 3 ) = 1.96 (Ans.) 

One could therefore expect to use this buffer to hold /?H's between 2 and 
4. Some uses of buffers in analysis are illustrated in the next chapter, 
where it will be demonstrated that certain precipitations are pH regulated 
and pH is in turn dictated by a buffer mixture. The precipitation reaction 
may be such that it would normally change the/?H and come to equilibrium 
with a low product yield, as Zn+ 2 + H 2 S = ZnS + 2H + . A basic buffer, 
by reacting with H 4 , promotes the forward reaction, since by Le Chatelier's 
principle, if the [H+] is kept low the equilibrium point is displaced to the 

Adjustment to Given pH 

By way of a corollary to the last paragraph, if a solution /?H is selected, 
one may predict which buffer solutions give about that pH and what 
specific mixture should be prepared. In practice one uses the approximate 
methods already outlined for the calculation, mixes the solution, measures 
the/>H with a/?H meter, and adjusts the/?H as needed with small additions 
of acid and base. 

Example 7-9. How many g of NH 4 C1 (mol. wt 53.5) are needed in a liter 
of 0.20 M NH 4 OH to give a soln. of/?H 9.0? 
The concn. of common ion NHj" will be about the same as the salt concn. 


and is the unknown in this problem. If the pH = 9.0, [H+] = 10~ 9 A/, and 
[OH~] = 10- 5 M. At equil., [NH 4 OH] ^ 0.20 A/, and K n = 1.8 x 10~ 5 
= [NH 4 4 ](10- 5 )/0.2, from which [NH^] = 0.36 A/. The g of salt needed per 
liter are (0.36)(53.5) = 19.3. (Ans.) 

A quick method of selection of the logical weak acid or weak base to use 
in a buffer for a certain pH is this: 

K A ^ the desired [H+], and K n ^ the desired [OH~] (7-17) 

The ratio weak electrolyte to salt is then ^1 and effective buffering action 
is assured. 

Some Buffer Mixtures 

Many buffer combinations are known and used. They may be made 
from salts and weak acids and bases as already illustrated, or simply by 
reaction between acids and bases in which at least one weak electrolyte 
is derived. Thus, upon mixing such rcactants as NaOH + H 3 PO 4 , 
KOH + H 3 BO 3 , Na 2 B 4 O 7 + HC1, NaOH + NaH 2 PO 4 , KH 2 PO 4 
+ Na 2 HPO 4 , Na 2 CO 3 + NaHCO 3 , H 2 C 2 O 4 + KOH, etc., buffer solutions 
are established. Calculations involving polyfunctional ions like H^Oj, 
HPO~ 2 , PO~ 3 , etc., are considered in quantitative analysis courses and will 
not be treated in much detail here. Approximations in these systems are 
usually justified and simplify otherwise lengthy calculations involving 
several simultaneous equations. 

Polyprotic Weak Acids 

Polyionic acids and bases introduce a complicating feature into problems. 
Such substances ionize in steps and an ionization constant can be found 
for each step. The first ionization always proceeds to a greater degree 
than the second and the second greater than the third, so the ion constants 
for the consecutive steps are in the order K > K 2 > K 3 . Not infrequently 
the latter dissociations are so weak that only the first is important to the 
problem. The reason for the decrease in acid strength is that with one 
less H+, the negative ion's attraction for the remaining hydrogen is greater 
since it is concentrated on fewer of them. A secondary cause is the com- 
mon ion effect, the H 4 now in solution acts to repress further dissociation. 
For a number of diprotic (also called dibasic) acids, the ratio of KJK 2 is 
approximately 1/10~ 5 or 1/10~ 6 , and for several triprotic acids, the ratio of 
is about 1/10~ 5 /10~ 10 . Appropriate examples are sulfurous acid, 

H 2 SO 3 = H+ + HSOg" ATi = 1.25 x 10~ 2 

HSOj = H+ + SOg 2 K 2 = 5.6 x 10~ 8 


and arsenic acid, 

H 3 AsO 4 = H+ + H 2 AsOj- K = 2.5 x 10~ 4 

H 2 AsOj- = H+ + HAsO- 2 K 2 = 5.6 X 10~ 8 

HAsO^ 2 = H+ + AsOj 3 K z = 3.0 x 10~ 13 

H 2 S 

In qualitative analysis the most important weak polybasic acids are 
hydrosulfuric, H 2 S, and carbonic, H 2 CO 3 , the former being of particular 
interest because it is the precipitating agent for cation groups 2 and 3. 
Its dissociations and constants (see reference 11, p. 126) are 

H 2 S = H+ + HS- (7-18) 

K^ = [H+][HS-]/[H 2 S] = 1.0 x 10~ 7 (7-19) 

HS- = H+ + S- 2 (7-20) 

#2 = [H + ][S~ 2 ]/[HS-] = 1.3 X 10~ 13 (7-21) 

Since K^ is roughly 10 6 times larger than K 2 , almost all the [H+] comes 
from equation 7-18 and further, since so little HS~ reacts in equation 
7-20 [H+] ^ [HS-]. It follows that if the latter concentrations are 
calculated as shown in equation 7-19 and substituted in equation 7-21, 
they will cancel, or [S~ 2 ] ^ K 2 . When K ^> K 2 for any weak diprotic acid, 
the molar concentration of the divalent ion has about the same magnitude 
as K 2 . 
For calculations involving the overall ionization, 

H 2 S = 2H+ + S- 2 

an equilibrium statement is obtained by multiplying equation 7-19 by 

K^Ki = K= [H+] 2 [S- 2 ]/[H 2 S] = 1.3 x 10- 20 (7-22) 

This is further simplified by learning that at room conditions of tempera- 
ture and pressure, the solubility of H 2 S gas in H 2 O is such that the solution 
is about 0.1 M in H 2 S. For problems in which saturated H 2 S solutions 
under laboratory conditions are described or inferred, one may use 

[H+] a [S- 2 ] = 1.3 x 10- 21 (7-23) 


[H+] = (1.3 X 10- 21 /[S- 2 ])* (7-24) 

It is evident that the solution pH will have a profound effect upon the 
[S~ 2 ] and precipitations using sulfide ion will need pH regulation. If 
too little S~ 2 is present (pH too low), metallic sulfides may not form or 


will be incompletely precipitated. If pU is too high [S~ 2 ] will be com- 
paratively large and some metals which one wants to retain in solution 
may precipitate. 

Example 7-10. Calculate the [H+] in satd. H 2 S soln. 

Since [S~ 2 ]c*K 2 = 1.3 x 10~ 13 , substitution in equation 7-24 of [S~ 2 ] 
should give us [H+]; [H+] = (1.3 x 10-~ 21 /1.3 x 10~ 13 )* 1.0 x 10~ 4 Af H+ 

One could have obtained the same result via equation 7-19, since if [H+] 
^ [HS~], then [H+J ^ (*i - [H 2 S])*. The student should verify this. 

Example 7-11. Calculate the [S~ 2 } in satd. H 2 S and compare it to the [S~ 2 ] 
in satd. H 2 S contg. enough HC1 to make the [H+] 2 M. 

Since K 2 ^ [S~ 2 ] = 1.3 x 10~ 13 Af, the first part is solved by inspection. 

Substitution of 2 M for [H f ] in equation 7-23 gives [S~ 2 J in the more acidic 
soln. ^ 3.3 x 10~ 22 M. (Arts.) The latter figure is about 10" 9 times smaller 
and illustrates the ease with which the concn. of the divalent ion of the weak 
acid can be changed with changes in /?H. 


F 3 

The carbonate ion is the precipitating agent for cation group 4 but the 
acid H 2 CO 3 is not as important to us as H 2 S. As one could predict from 
the foregoing, two dissociations are possible: 

H 2 CO 3 = H+ + HCO 3 - 

*i = [H'][HC03]/[H 2 C0 3 ] = 4.16 x 10~ 7 
HCO- = H* +0)3-* 

^2 = [H^ICO^/tHCOj] - 4.84 x 10~" 
K& = K= [H+]2[C03 2 ]/[H 2 C0 3 ] - 2.01 x 1Q- 17 

A water solution saturated with CO 2 at room conditions is about 0.034 M 
in H 2 CO 3 . For such solutions one may incorporate the saturation concen- 
tration in the last equation and get 

[H+RCOg- 2 ] = 6.85 x 10~ (7-25) 


[H+] = (6.85 x 10-"Y[C0 3 2 ])* 

As approximated in the H 2 S case, we may say [H f ] ^ [HCOg] and 
[COj 2 ] ^ K 2 . These are the identities most often used in carbonate 

Carbonates are in general much more soluble than the corresponding 
sulfides and hence a greater [CO- 2 ] than [S~ 2 ] is needed for their precipita- 
tion. From equation 7-25 one should observe that carbonates are 


formed at high pR and dissolved at low pH. This will be considered 
further in the next chapter. 

Strong Polyprotic Acids 

If one is dealing with stronger acids than H 2 S and H 2 CO 3 , wherein 
the several ionization steps make significant contribution to the final 
total ionic population, then all the ionization constants become important 
to the problem. The most common example of this is sulfuric acid. It 
is assumed that in dilute solution, dissociation of one H+ is complete: 

H 2 S0 4 -> H+ + HSOj- 

The second dissociation, however, is weak enough so that equilibrium 
principles may be applied to it, yet not so weak that it can be ignored as a 
source of H 4 : 

HSO- = FT* + SO" 2 

K 2 = [H'][S0 4 2 ]/[HSO-] = 1.26 x 10~ 2 (7-26) 

Example 7 12. Calculate the approx. concn. of each ion in 0.10 M H 2 SO 4 
and the approx. /?H. 

Since the first dissoc. step is assumed to be complete, momentarily [H f ] 
= [HSO^] = 0.10 M. Bisulfate ion then undergoes ioniz. giving more H+ 
and sulfate ion. If we let r be the molar amt. dissoc. at equil. 

HS0 4 ^ H+ + SOJ 2 (7-27) 

(O.IO-.T) (O.I 

Substituting in equation 7-26 gives 

1.26 x 10~ 2 = (0.10 + x)(x)/(OAO - sr) 

From the quadratic, x r^ 0.01 M. 
From equation 7-27 one gets the following for the ionic concns.: 

[SCV] ^ 0.01 M, [HS0 4 ] ^ 0.099 A/, and [H+] ^ 0.1 1 M 
corresponding to a pH of 0.96. (Ans.) 

Polyhydroxy Bases 

For most purposes in qualitative analysis, calculations concerning 
equilibrium constants of strong polyhydroxy bases such as Ca(OH) 2 and 
Ba(OH) 2 are of little interest and are handled as we handled the sulfuric 
acid case. Most other polyhydroxy bases are only sparingly soluble in 
water, such as Fe(OH) 3 , A1(OH) 3 , etc., and will be considered in the next 
chapter, as their uniqueness lies in their insolubility rather than their base 
strength. Weak soluble polybases are almost all organic nitrogen com- 
pounds and the calculations in which they appear are similar to those with 
H 2 S and H 2 CO 3 . 


Benzidine, abbreviated B(NH 3 OH) 2l is representative of this class and 
its ionization resembles that of ammonium hydroxide: 

HOH 3 N-f V-f V-NH,OH 

= HOH 8 N-/~\-/~~\-NHJ + OH- 

\- - / V- ,-/ 

= [BNH 3 OHNH+][OH-]/[B(NH 3 OH) 2 ] = 9.3 x 1<H 



= HJN-/~~Y-/~~-NHJ + OH- 

[B(NH 3 ) a ][OH-]/[BNH 3 NH 3 OH+] = 5.6 x 



Indicators for /?H determination are prganic weak acids and bases that 
change color by an alteration of structure upon accepting or donating H+. 
By knowing their ionization constants one can calculate such things as 
their pH range. Consider some indicator HIn, a weak acid: 

HIn = H+ + In- 
K A - [H'][Iir]/[HIn] or [H+] = AT.|([HIn]/[In"]) 

The colored forms are HIn (color at lower /?H) and In~ (color at higher 
/?H). The intensity of color observed is proportional to the concentration 
of these substances and, when they are in equal concentration, they cancel 
from the equation above, and [H+] = K A . The corresponding pH is the 
midpoint of the indicator's pH range. The observed color is a mixture of 
the colors of the two forms. The eye cannot distinguish well between the 
two colors unless one colored form has a concentration about ten times 
the other form, hence the range for indicators of this type is given by, 
[H+] = ^-(10/1) and [H+] = ^-(1/10), or pH = pK A I. 



1. Explain in a paragraph, using complete sentences, equations, diagrams, 
etc. as needed, how the system, NH 4 OH + NH 4 NO 3 resists a pH change when 
small amounts of strong acid or base are added. 

2. For ratios 1/10 and 10/1 of weak electrolyte to salt, calculate the/>H range 
for the buffer in problem 1. What assumptions are made in your method? 

3. At 1 8 C and a potential of 1 10 volts, 0.041 amperes flow through a resistance 
cell (with a cell constant of 1.00) dipping into a 0.002 M solution of weak acid 
HA. At infinite dilution, the equivalent conductance of A~ is 35.0. Calculate 

(a) specific resistance and conductance of the solution (b) ml volume necessary 
to contain 1 g equiv. wt of HA at 2 x 10~ 3 M (c) equivalent conductance of 
solution (d) equivalent conductance of HA at infinite dilution and (e) % ioniza- 
tion of HA at 2 x 10~ 3 M. [Ans. (a) 2.69 x 10 3 , 3.73 x 10~ 4 (b) 5 x 10 5 
(c) 186 (d) 350, and (e) 53.4.] See methods, p. 90-91 Chapter 5. 

4. Use (a) a more exact and (b) an approximate method for calculating the % 
ionization in 0.005 M HAc. Compare the results by giving a % deviation in 

(b) from the other answer. Explain why the answers are different. 

5. HX is a weak acid, K A = 6.4 x 10~ 5 . At what concentration is HX 
2.5 % dissociated ? (Ans. 0. 1 6 M). 

6. A solution is prepared by mixing 100 ml of 0.2 M HNO 2 with 100 ml of 
0.1 M NaOH. Calculate the approximate pH of the mixture and list each 
assumption you made in working the problem. 

7. To 100 ml of a solution 0.2 M in HAc and 0.1 M in KAc is added 20 ml 
of 0.02 M NaOH. Calculate the initial /?H, the final pH and the % change 
in pR. 

8. Calculate the original and final /?H: 

(a) 100 ml of 0.04 M HAc diluted to 4 liters 

(b) 100 ml of 0.04 M HBr diluted to 4 liters 

Explain why these solutions change pH by different amounts, and why the 
original pH's are not identical, whereas the original concentrations of acids are. 

9. Calculate the approximate concentration of each ion, including OH~ in: 

(a) 0.01 M H 2 SO 4 

(b) 100 ml of 0.01 M H 3 AsO 4 

10. At 18 C, A for HC1 = 380, for NaAc = 78, and for NaCl = 109. From 
these alone calc. A for HAc. (Ans. 349). 

11. One hundred ml of 0.10 M HAc are mixed with 100 ml of 0.04 M NaOH 
and 100 ml of H 2 O. Calculate the final [H+], [OH~], /?H, and pOH. 

12. At 18 C, A for NaCl = 109. For other dilutions of NaCl, A. 01 y = 102.8 
andA.uy = 92.5. Calculate the apparent % ionization and AT at each dilution. 
Explain why K changes with concentration. 

13. Calculate the degree of ionization in 10~ 3 M HCN. At what concentra- 
tion is the ionization 10%? 

14. Equation 7-16 was developed for an acid-type buffer. Develop a general 
formula for finding pH of a base-type buffer. 


15. What concentration of HAc has a pH of 5.25? What concentration of 

16. For a weak acid HQ, if [B] = [HQ] show that approximately, [H+] 2 /[B] 

17. From the result in 16, show that [H+] = ([B]^)*, and derive a general 
formula for pH from that. For what cases could it be used? 

18. Four mmoles of H(CO 2 H), formic acid, and 5.5 mmoles of Na(CO 2 H) 
are dissolved in H 2 O to make 50 ml of solution. Calculate the pH. 

19. How much solid NaAc must be added to 3 ml of 0.28 M HC1 so that 
when the solution is saturated with H 2 S, [S~ 2 ] = 10~ 14 Ml 

20. To what volume must 10 ml of 0.3 M HAc be diluted to give a solution 
of /?H 4.32 ? (Ans. 23.4 liters.) 

21. An unknown weak, monoprotic acid has apH of 4.76 in 0.01 M solution. 
Calculate K A and from the table of ionization constants, determine which acid 
it might be. (Ans. HC1O, experimental K t = 3.03 x 10~ 8 ). 

22. One has 100 ml of 0.02 M NH 4 OH. Which of the following would be 
most effective in lowering the pH : (a) 100 ml of 0.002 M HC1 (b) 0.2 moles of 
NH 4 C1 (c) 900 ml of H 2 O? Explain with calculations. 

23. A solution of HAc is 0.074 M and has an equivalent conductance at 25 C 
of 6.09. With the aid of Table 5-2 calculate the pH of the solution. 

24. Will 50 ml of 0.2 M HAc require more, less, or the same amount of 
NaOH to react with it stoichiometrically than 100 ml of 0.1 M HNO 3 ? Explain. 

25. One has a 0.1 M solution of the theoretical weak acid H 2 Q, K Y = 10~ 3 
and K 2 = 10~ 9 . Calculate the approximate [Q~ 2 ]. Explain any shortcuts. 
(Ans. 10- 9 M) 

26. From the data in 25, (a) explain why the answer was independent of the 
solution concentration and (b) calculate the value of [H+]*[CT 2 ]/[H 2 Q]. (Ans. 
(b) 10- 12 ) 

27. The actual pH of 0.10 M H 2 SO 4 is 1.1. Why is the pH calculated in 
example 7-12 not the same? 

28. (a) One has two HC1 solutions, one of pH 3, the other of pH 2. If 10 ml 
of each are mixed, what is the resulting ^H? 

(b) A hundred ml of 0.1 M HC1 is mixed with 100 ml of 0.1 M HAc. Calculate 
the pH of the mixture. Explain any assumptions. 

29. (a) Describe an experiment by which one could determine the ionization 
constant of a monoprotic weak acid. 

(b) Describe another experiment whereby one could determine which of 
two weak acids is the weaker, given 0.1 M water solutions of each. 

30. Arrange these in order of increasing acidity: H 2 O, 0.2 M HAc, 10~ 5 Af 
HC1, 2 M NHJ in 10~ 3 M NH 4 OH, 0.1 M H 3 PO 4 . 

31. Jackson Slipshod knows that 0.1 M HAc is 1.34% dissociated. To 
1 liter of such solution he adds 0.1 mole of solid NaAc. He wants to calculate 
a value for K A from the known concentrations and reasons that [H+] =* (0.1) 
(0.0134), [Ac~] = (0.1)(0.0134) + 0.1, and [HAc] = (0.1)(0.9866). Putting 
these figures into K A = [H+][Ac~]/[HAc] he obtains an answer larger than the 
accepted K A . What is wrong with the Slipshod method? 


32. A certain indicator of the type HIn has a blue color at/>H 4.5 and yellow 
color at/?H 6.5. Find the ionization constant of the indicator. 

33. Under what conditions may one write equation 7-1 as 

HQ ^ H+ + Q- 

Af ocAf ocM 

34. Equations 7-10 and 7-12 were developed for a weak acid. Develop 
analogous equations for a weak base and explain under what circumstances the 
equations can validly be utilized. 

35. An indicator is a weak base, InOH, K u = 1 .5 X 10~ 6 . (a) Give formulas 
for the colored forms. (/>) Calculate its pH range, (c) Name an acid and a 
base that could be properly titrated using the indicator. 

36. With reference to the section on strengths of oxygen acids, "guessimate" 
K A for the following and check your answers with the appropriate table: HBrO, 
H 2 C 2 O 4 , HC10, H 2 S0 3 . 


1. B. Naiman, J. Chem. Educ., 25, 454 (1948). (Bronsted concepts) 

2. K. J. Radimer, J. Chem. Educ., 27, 251 (1950). (Equilibrium problems) 

3. D. D. DeFord, J. Chem. Educ., 27, 554 (1950). (Bronsted calculations) 

4. R. N. Boyd, /. Chem. Educ., 29, 198 (1952). (Equilibrium problems) 

5. A. J. McBay, /. Chem. Educ., 29, 526 (1952). (Approximate /?H) 

6. B. Park, J. Chem. Educ., 30, 257 (1953). (Weak acids) 

7. J. J. Fritz, J. Chem. Educ., 30, 442 (1953). (Equilibrium calculations) 

8. J. W. Drenan, J. Chem. Educ., 32, 36 (1955). (Calculation of [H ;J O+]) 

9. D. Davidson, J. Chem. Educ., 32, 550 (1955). (Equilibrium and amphotensm) 

10. G. Gorin, /. Chem. Educ., 33, 318 (1956). (Indicator constants) 

11. J. W. Kury, A. J. Zielen and W. M. Latimer, /. Electrochem. Soc., 100, 468 (1953). 
(H 2 S equilibrium constants) 






Water is an excellent solvent, as described in Chapter 3. When it 
dissolves substances, energy is released as ions become hydrated and this 
energy is used to overcome interionic forces in salt crystals. If the hydra- 
lion energy is greater than the lattice energy, the solid dissolves; if not, low 
solubility is the result. A given amount of water at a fixed temperature 
will dissolve a definite amount of solid when excess solute is shaken with it. 
The reason variable solubility is not noted under these conditions is that 
the solubility process is reversible and is ultimately balanced by a counter 
reaction in which ions are reprecipitated on the surface of the excess solute. 
If the ionic concentrations in solution are changed say by dilution, the 
compound will dissolve more and the system will re-establish the equili- 
brium concentrations again. It is found that many of the ideas of general 
homogeneous system chemical equilibrium can be related to heterogeneous 
systems when the solvent is water and the solute is a sparingly soluble 
electrolyte. It is the purpose of the following paragraphs to apply this 
quantitatively to some problems stemming from the laboratory part of 
the course. 

The Solubility Product Constant 

When one causes a precipitate like AgBr to form in the laboratory, 
he may believe that if stoichiometric quantities were used all Ag+ and Br~ 
have been removed from solution. This is not the case for any precipita- 
tion. The solution remains saturated, though sparsely populated, with 



ions, and equilibrium is set up between them and ions on the precipitate's 
surface. Simultaneous dissolution and reprecipitation are pictured as 
going on at the surface and the ideas of Chapter 6 are logically extended to 
include this reversible process: 

AgBr = Ag+ + Br- 
The equilibrium statement will be 

K = [Ag+][Br-]/[AgBr], or X[AgBr] = [Ag+][Br-] 

We will consider all solids to have a constant concentration of unity so such 
terms as AJAgBr] are constants, and will henceforth be designated K 8P , 
or the solubility product constant: 

K S r = [Ag + ][Br-] (8-1) 

As with equilibrium applications in the two previous chapters, if several 
similar ions are involved as 

Bi 2 S 3 = 2Bi+ 3 + 3S~ 2 
then the coefficients are used as powers in the K SP expression: 

K ap = [Bi* 3 ] 2 [S- 2 ] 3 (8-2) 

If the several ions produced on dissociation are all different, then coeffi- 
cients do not appear, as 

Mg(NH 4 )P0 4 = Mg+ 2 + NH+ + PO- 3 

K SP = [Mg+ 2 ] [NH+] [P0 r 3 ] (8-3) 

At a given temperature and in the absence of substantial concentrations of 
other ions, K SP is constant for a given sparingly soluble salt and is equal to 
the product oj the molar concentrations of the ions each raised to a power 
which is its coefficient in the equation, showing the compound's dissociation 
into ions. 

Factors Affecting Solubility 

A number of commonly encountered influences affect solubility and are 
of concern to the analyst because of the many times he must dissolve 
precipitates or must precipitate a substance quantitatively in a state of 
high purity and suitable for centrifuging or filtering. 

1. Temperature. In most cases, the solubility of salts (and hence K SP ) 
increases with increasing temperature, though some salts like NaCl show 
little solubility change and a few, such as Ce(SO 4 ) 2 , show decreasing 


solubility. If the heat of solution is negative (the solution becomes cooler 
as salt is dissolved), the compound's solubility increases as the temperature 
is raised, and, if the heat of solution is positive, then raising the tempera- 
ture decreases the compound's solubility. This is another illustration of 
Le Chatelier's principle. If some solid salt having a positive heat of solu- 
tion is in equilibrium with its saturated solution, and the system's tempera- 
ture is lowered, more solid goes into solution evolving heat to restore the 
original temperature. By similar reasoning one can deduce the opposite 
is true for compounds having negative heats of solution. Tables in this 
book will give K S1 , values at room temperature. 

2. Indifferent ions. The AT >s/ ,'s in the table were determined for 
saturated solutions of single compounds in pure water. If soluble salts, 
furnishing indifferent or uncommon ions, are present they change the 
ionic character of the solution and increase the solubility of slightly 
soluble salts by the salt effect (see Chapter 7). A simple explanation for 
this is that the concentrations of ions of the low-solubility salt in solution 
are effectively diminished due to interionic attractions that slow their 
movement. If such is the case then the rate of reprecipitation falls below 
the rate of solution. Solubility product constants will thus change 
slightly, but this refinement is omitted in beginning work and we shall use 
the K Kl * values as given, and molar concentrations, in calculations 
concerning them. 

3. Different solvents. Discussion, application, and data will be 
limited to water as a solvent but different solvents obviously change the 
solubility of a given substance. Organic solvents like alcohol are fre- 
quently added to water solutions of salts to precipitate the salts, and the 
reverse is also done salts like NaCl being added to aqueous solutions of 
organic compounds (as soap) to precipitate them a process known as 
salting out. 

An application of the use of a nonaqueous solvent can be made in the 
separation of SrCrO 4 and CaCrO 4 in the cation group 4 procedure. Both 
chromates are fairly soluble in water and both are insoluble in alcohol. 
By making the proper mixture of these solvents, however, one is able to 
precipitate only the less soluble strontium chromate and leave the calcium 
compound in solution. 

4. Complex formation. It is clear that if ions are present which will 
form complexes with ions from the low-solubility salt, the solubility of the 
latter may become great. Thus the solubility of AgBr in H 2 O is low, 
K SP = 3.3 x 10~ 13 , but, if NH 4 OH is present, the solubility increases 
markedly due to formation of Ag(NH 3 )+, which is moderately stable. 

5. Particle size. Small particles of a given precipitate are more soluble 
than large particles, in some cases as much as a thousand times more 


soluble.* The reason for this behavior is that the small particles have a 
higher surface tension than the larger crystals and the tendency is toward 
reduction of this energy by reorganization to produce larger crystals of 
smaller surface area. Crystalline substances such as BaSO 4 and CaC 2 O 4 
when precipitated rapidly contain many small crystals which are difficult 
to separate from the solution but which will increase in size if allowed to 
age or digest, that is, come to equilibrium with solution and the rest of the 
precipitate in such a way as to reach a fairly coarse average particle size. 

Ca(N0 3 ) 2 = high ionic strength 

HBr = common ion, then complexing, AgBr 2 

Concentration of added substance 

FIG. 8-1. Schematic review of important factors affecting the solubility of a salt. 
Point x represents the saturated aqueous AgBr solution. 

Solubility product constants are given only for this stable condition. 
Amorphous precipitates like hydrated oxides have no definite orientation 
as precipitated, and on standing may change solubility considerably, 
commonly getting less soluble with time. 

6. Common ions. The effect in the solution of ions that are common 
to ones in the slightly soluble salt will at first decrease the solubility of the 
latter as explained in the last two chapters. Thus, in precipitating a 
substance from solution, a little excess precipitating agent is used for its 
common ion effect. This shifts the equilibrium by the mass action prin- 
ciple to produce a maximum yield of desired precipitate. A large excess is 

* The relationship is given by (RT/mol. wt) In S r /S 2a/dr, in which R is the gas 
constant, T absolute temperature, *S' r the solubility of fine particles of radius r, S the 
solubility for coarse equilibrium particles, a the surface tension, and d the density of the 
particles. The formula is derived by thermodynamics. 


avoided, however, because of the salt effect or possibility of complex 
formation, as 

HgI 2 + 21- = 

7. Hydrolysis. If the cation of the sparingly soluble salt is derived 
from a weak base and/or the anion is derived from a weak acid, either or 
both ions may react with water (in a manner detailed in Chapter 10) to 
give hydrolysis products and a resulting increase in solubility. For 
example CaC 2 O 4 is expected to hydrolyze some due to the weakness of 
oxalic acid : 

CaC 2 4 + H 2 = Ca+ 2 + C 2 O" 2 + H 2 O - CV 2 + HC 2 Oj + OH~ 

The result is that CaC 2 O 4 is more soluble than if hydrolysis did not occur. 
The effect can be quite appreciable in these very dilute solutions, particu- 
larly with sulfides since the K A for H 2 S is so small. 

Determination of Solubility Product Constants 

Af w /s are calculated from solubility data gathered by several methods, 
some of which are explained below. 

1. Electrical conductivity. As with weak acids and bases, one can 
use electrical conductivity of the solution to determine the ionic concentra- 
tions in it. In Chapter 5 the formula A = /<:F was developed in which A 
is the equivalent conductance, K the specific conductance, and V the 
cc volume needed to contain 1 g equivalent weight of solute. If we let 
S be the solubility of the slightly soluble salt expressed in gram equivalent 
weights per cubic centimeter, then K, which was the cc per g equivalent 
weight, is its reciprocal, or A = K/S. Since the solution is so dilute, A 
approaches its maximum value A , which is the equivalent conductance at 
infinite dilution. One can obtain its value from addition of ionic con- 
ductances given in Table 5-2. The formula then becomes 5 = *c/A . 
Since a more useful solubility than gram equivalent/cubic centimeter would 
be gram equivalent/liter, let 5' be the solubility in these concentration 
terms. Then 5/1000 = 5', and the formula now derived for greatest 
usefulness becomes 

S' = lOOO/c/A (8-4) 

Example 81. The sp. cond. of AgBr at 18 C is 1.23 x 10 
and the sp. cond. of H 2 O at that temp, is 1.16 x 10~ 6 ohm ^cm" 1 (a correction 
that is significant at these dilns.). For AgBr alone, K = 0.07 x 10' 6 . Cal- 
culate the KN/> of AgBr from that. 

From Table 5-2, A Ag , + A 1Jr . = 121.6 = A for AgBr, and ,S" = (1000) 
(7 x 10~ 8 )/121.6 = 5.76 x 10" 7 g equiv./liter. This figure will be the same as 
the molar solub., since the equiv. and mol. wts of AgBr are equal. Dissociation 


being approx. 100% at such low concn., [Ag+] = 5.76 x 10~ 7 M = [Br~] and, 
K SP = [Ag+][Br-] = (5.76 x 10- 7 ) 2 = 3.32 x 10~ 13 (Ans.) 

2. Electrochemical cells. By measuring the voltage of properly 
designed cells, one is able to relate it to the concentration of ions in a 
saturated solution of a slightly soluble salt and thus to calculate K KP . 
(The calculation is similar to that of Example 11-5.) 

3. Radioactive tracers. The number of curies* of radioactivity, 
determined by a counter, emitted by a gram of radioactive material is 
defined as the specific activity of the material. Because the number of 
nuclear disintegrations that radio elements undergo per second is very 
large, only traces of the elements or their ions need be present to give a 
detectable count. Various elements have been easily determined in 
microgramf per liter amounts. Since the count is related directly to the 
amount present, if one knows the specific activity of the element and the 
activity in a saturated solution of a slightly soluble salt containing it, he 
is able to determine the solubility and K s[ >. Solutions may also be 
evaporated and the activity of the residue checked. 

4. Simple evaporations. Where salts are soluble enough to yield a 
weighable residue upon evaporation of a reasonable volume of their 
saturated solutions, this is the most direct method of determining solubility 
product constants. See problem 2 and Example 8-2. 

Example 8-2. It is found that a liter of satd. BaSO 4 (mol. wt 233) at 18 C 
leaves a residue of 2.3 mg of salt upon evap. Find the K$ P . 

The molar solub. is 2.3 x 10~ 3 g/233 ^ 9.9 x 10~ 6 moles/liter. The dissoc. 
is complete so this is also the molarity of Ba +2 and of SO^ 2 in equil. at 18 C 
with solid BaSO 4 and 

[Ba+ 2 ][S04- 2 ] = (9.9 x 10" 6 ) 2 = K SP = 9.8 x 10' 11 (Ans.) 

5. Turbidimetry and nephelometry. The first of these methods 
concerns the detection of solubility limits. One causes one dilute solution 
to react with another until turbidity due to the appearance of a precipitate 
is just noticed. This is the saturated solution, and K SP is calculated from 
knowing the concentrations. Turbidity is determined by shining a beam 
of light through the solution; only when the first very fine solid particles 
are present will the beam be scattered by particles and appear to extend 
through the suspension as a shaft of light. 

The second method involves shaking a known quantity of solid solute 
with an increasing volume of water and measuring the light scattering 
due to undissolved suspended solid. This is done, as above, by means of 

* 1 curie = 3.7 x 10 10 disintegrations/sec. 
t 1 microgram, ftg = 10~ 6 g. 


an optical instrument known as a nephelometer. The dilution beyond 
which solvent addition results in no further reduction in light scattering 
is the saturated solution and K SP follows by calculation. 

6. Ultramicroscopy. In this method, one studies successive dilutions 
of solid in water using the ultramicroscope to determine which dilution 
just fails to show any solid phase. This is the saturated solution, and 
again A^ is calculated from the known concentrations. 

7. Colorimetry. Sometimes the substance being studied lends itself 
to a colorimetric measurement that enables one to determine equilibrium 
concentrations of ions in the saturated solution. This is illustrated in 
special experiment 3 wherein the K sr of CaF 2 is found with the aid of a 
sensitive organic reagent for fluoride. Many experiments can be designed 
using that general principle see problems 3 and 4, for example. 

K SP and Solubility 

K NP values can be misleading in the sense that they do not express 
solubility directly, although they are related to it. The following example 
illustrates both this and the method for determining solubilities from the 
KM* table. 

Example 8-3. Cul and Ag 2 CrO 4 both have about the same K SP value, 
1.1 x 10~ 12 . Calculate the molar and g/liter solub. for both. 

Taking Cul first, if we let x be the molar concn. of Cu f in the satd. soln., then 
.r is also the concn. of I", since the two ions are dissoc. in equal numbers. 

[Cu+][I-] = KBP = (x)(x) = 1.1 x 10~ 12 = x 2 

and x = 1.05 x 10~ 6 M. This is also the number of moles of Cul which will 
dissolve to make a liter of satd. soln., so the molar solub. = 1.05 x 10~ 6 
moles/liter (Ans.), and the g/liter solub. is obtained by multiplying by the 
mol. wt, 190.5, giving 2.0 x 10~ 4 g/liter. (Ans.) 

If y is the molar concn. of CrO^ 2 in a satd. soln. of Ag 2 CrO 4 , then [Ag+] will 
be 2y since 2Ag + are present with each CrO 4 ~ 2 : 

Ag 2 Cr0 4 = 2Ag+ + CrOJ- 
2y v 

Substituting in the K SP statement and solving for y, 

[Ag+] 2 [Cr0 4 - 2 ] = K SP = (2t/) 2 (y) = 1.1 x 10" 12 = 4y 3 
y = (1.1 x 10- 12 /4) 1/3 = (275 x 10~ 15 ) 1/3 = 6.5 x 10~ 5 M 

Since y moles of CrO^ 2 come from y moles of Ag 2 CrO 4 , the molar solub. of this 
salt is 6.5 x 10~ 5 moles/liter (Ans.), and the g/liter solub. is (6.5 x 10- 5 )(331.8) 
^ 2.2 x 10~ 2 (Ans.). 

Thus though the K SP 9 s were about the same, the squared term in the 
Ag 2 CrO 4 case and the difference in molecular weights of the two salts led 
to the fact that the solubilities are not identical. 


The opposite case is true also; if one has two salts such as AB and 
C 3 D which have the same solubility in grams per liter, one can see that 
different molecular weights will lead to different molar solubilities and the 
cubed term will give a considerably different K$ P . 

Conditions for Precipitation 

A common problem in analytical chemistry is that of discovering condi- 
tions of concentration under which precipitation will or will not occur. 
If such knowledge is available, schemes of gravimetric separations follow. 
Since the K sr 's are calculated from data on saturated solutions it should 
be clear that any greater concentration of ions involved than those which 
give the K SP value by proper manipulation, are in excess of equilibrium 
concentrations and precipitation will take place (assuming supersaturation 
does not occur). This type of problem reduces itself to a comparison 
between the magnitude of the K S1 , and the product of the ionic concentra- 
tions (properly handled algebraically to match the K SP definition for the 
compound being studied, cubing, squaring, etc., as needed). The product 
of ionic concentrations so manipulated is called the ion product, LP. 
Three cases arise: 

LP. > K sl >; precipitation takes place until I. P. = K HP . 
I. P. = K KP ; no apparent reaction; solution is saturated. 
LP. < KN P ; no precipitation; precipitate already present dissolves 
until LP. = KB?. 

It should now be added that the precipitate must not only be theoreti- 
cally capable of appearance, but sufficient quantity must be present for 
the observer to see. If one is using a milliliter of test solution, he is 
capable of detecting no less than between 0.1 and 0.01 mg of average 
precipitate suspended in it. If precipitates have an average molecular 
weight of about 150, this means that the concentration of precipitate to be 
detected must at least be in the approximate range 7 x 10~ 4 7 X 10~ 5 M. 

Example 8-4. A certain tap water is known to contain 300 parts per million 
(ppm = mg/liter for dil. aq. soln.) sodium chloride. If to 50 ml of this are 
added 50 ml of 0.02 M AgNO 3 , will AgCl pptn. take place? 

All one needs do is to compare the LP., [Ag+][Cl~] in the mixt. to the K K p, 
whose value from the table is 2.8 x 10~ 10 . If LP. > K SP , a ppt. results. The 
NaCl concn. in the tap water is 3.00 x 10~ 1 g/liter divided by the mol. wt, 
58.5, or 5.13 x 10 3 M. Since each soln. dilutes the other in mixing, then, 
momentarily, prior to reaction the NaCl concn. is 2.56 x 10~ 3 Af and the 
AgNO 3 is 10~ 2 M. Therefore [Ag+] = 10 2 M and [Cl~] = 2.56 x 10~ 3 M, 
and the LP . [Ag+][Cl ] = 2.56 x 10' 5 , which is larger than the K SP so AgCl 
does ppt. AgCl continues to form until LP. = K^ P . (Ans.) One can further 


calc. that the amount of ppt. formed is sufficient to be visually detected. (How 
is this done?) 

Fractional Precipitation 

If a solution contains two or more ions that can form slightly soluble 
compounds upon addition of some oppositely charged ion, the precipitates 
will form in the order which their K sl Ss are exceeded. If one is to effect 
good separations in this way, the quantity of the first precipitating com- 
pound still remaining unprecipitated must be very small when the I. P. of 
the next precipitating compound just exceeds its K sr . Naturally the 
greater the difference in solubilities of the two or more compounds to be 
precipitated the more readily and clean cut may the separation be made, 
For practical purposes, the equilibrium ratio between two ions to be 
separated in this way should be at least 10 5 or 10 6 to 1, a ratio one can 
determine with the aid of K S2 > data. The precipitation of a selected 
species from solution in this manner is called fractional precipitation or 
fractional crystallization, and is a principle used frequently in analytical 
chemistry. Two applications from this course follow. 

/. Group 2 and Group 3 Sulfide Separations. If one calculates 
the solubilities of the group 2 sulfides from K KP data and compares them 
with the solubilities of the group 3 sulfides, he finds the latter are consider- 
ably more soluble. Thus for a given [S~ 2 ], more group 3 metal ions are 
needed in solution for equilibrium than group 2 ions, hence through proper 
regulation of [S~ 2 ] it is possible to precipitate only group 2 sulfides and 
leave group 3 metals still in solution. When the former have been 
removed by centrifugation, the latter are precipitated using the same 
precipitant, H 2 S, but regulating conditions to produce more [S~ 2 ]. This 
is most easily done with/?H control since [H^] is related to [S~ 2 ] by the 
H 2 S equilibrium. A solved problem should clarify these generalizations. 

Example 8-5. Calculate the approx. mg of Cu+ 2 (typical group 2 ion) and 
mg of Fe+ 2 (typical group 3 ion) that could remain unpptd. in 1.0 ml of soln. 
that is 0.3 M in H+ and is kept satd. with H 2 S (i.e., is under group 2 pptn. 

From [H+] 2 [S- 2 ] = 1.3 x 10~ 21 for satd. H 2 S, one can determine the sulfide 
in equil. with 0.3 M H+ to be 1.4 x 1Q- 20 M. (See Chapter 7 on H 2 S.) This 
being the [S~ 2 ] available for pptn. it remains only to find from their respective 
solub. product constants the quantity of metallic ions that can be tolerated 
simultaneously in soln. 

For CuS, K sr = 4 x 10~ 36 = [Cu +2 ][S~ 2 ] and substituting 1.4 x 10" 20 M 
for [S~ 2 ] gives a value of 2.8 x 10~ 18 M which corresponds to 1.8 x 10~ 14 mg 
of Cu+ 2 left unpptd. in 1 ml of soln. (Ans.) By the same method, using 
4 x 10~ 17 for the K SP of FeS one calcs. that the equil. [Fe+ 2 ] is 2.8 x 10 3 A/, 
corresponding to 1.6 x 10 5 mg of Fe+ 2 /ml left unpptd. (Ans.) 


It is evident that essentially all Cir 12 precipitates and no Fe+ 2 can 
precipitate (regardless of how much was originally in solution), as the 
theoretical equilibrium [Fe+ 2 ] greatly exceeds the solubility of any ferrous 
salt. These being typical ions of the two analytical groups, one sees how 
effective and important fractional precipitation methods can be. 

The problem just solved was done in a very approximate way and a 
refinement should be noted now. Since the Cu +2 precipitation is found 
to be all but complete 

Cu f2 + H 2 S = CuS + 2H+ 

shows that as reaction proceeds, the /?H of the solution continues to drop. 
This increase in [H+] is considerable if a large amount of CuS precipitates, 
and since [H 4 ] is a major factor in sulfide precipitations it is advisable to 
work such problems allowing for [H+] change, as below. 

Example 8-6. One has a 0.3 M H+ soln. contg. 10 mg Cu+ 2 /ml. If one 
keeps 1 ml of this saturated with H 2 S, calculate the [Cu+ 2 ] in soln. at equil., 
taking into account the fact that the [H+] will increase. 

From Example 8-5 one may expect that almost all the copper will be pptd. 
as the sulfide. The original [Cu +2 ] is 10/63.5 = 0.158 A/, and since two H+ are 
formed for each Cu+ 2 pptd., the H+ increases from 0.3 M to 0.3 -f (2)(0.158) 
sz 0.62 M. From K A for satd. H 2 S, one calculates that 0.62 M H f depresses the 
ioniz. to the point where [S~ 2 ] is only ^ 3.4 x 10~ 21 M. From the K Kl > of 
CuS one then finds that the [Cu+ 2 ] in equil. with this amount of sulfide is 
^ 1.2 x 10~ 15 M. (Ans,) Due to the very small value of the K SP , the [Cu+ 2 ] 
is still minute, so in this case the additional [H 4 ] did not hinder the pptn. 

In another situation where different concentrations and a more soluble 
sulfide were considered, the more exact calculation might demonstrate 
that precipitation would never take place quantitatively. 

The next example shows another method for handling this type of 

Example 8-7. Calculate the ratio [Fe+ 2 ]/[Cu+ 2 ] that could exist in soln. 
when, as in Example 8-6, the [H + ] is 0.62 M and [S~ 2 ] is 3.4 x 1Q- 21 Show 
that the ratio is greater than 10 6 to 1, hence separation by fractional pptn. 
is feasible. 

If one simply multiplies each term of the desired ratio by [S~ 2 ], the expression 
becomes a ratio of K 8 p* of the two sulfides. These being given, the problem 
is solved in one step: 

HS^J/tS- 2 ] = [Fe+ 2 ][S- 2 ]/[Cu+ 2 ][S- 2 ] 

4 X 10~ 17 /4 x 10~ 36 
1 X 10 19 /1 (Ans.) 


The ratio of metal ions left in soln. being much greater than 10 e to 1, separation 
by fract. pptn. is feasible. (Ans.) 

Multiplication by factors like [S~ 2 ]/[S- 2 ] to simplify the computation by 
producing expressions whose values are known challenge the alertness of 
the student and allow for some quick and original solutions to problems. 

2. Mg+ 2 and Group 4 Separation. A further illustration in pH control, 
which regulates in turn the concentration of ions causing precipitation, is 
afforded in the case of Mg+ 2 , whose hydroxide precipitation is prevented in 
group 4, though the solution is basic with an NH 4 C1 NH 4 OH buffer. 

Example 8-8. One has 2.0 ml of soln. contg. K) mg of Mg (~ 0.21 M Mg+ 2 ), 
2.0 M NH/, and 0.30 M NH 4 OH. Show that Mg(OH) 2 cannot ppt. 

From the last two concns. and K tt for NH 4 OH, one obtains K B = 1.8 x 10~ 5 
= (2.0)[OH-]/(0.30), from which [OR-] = 2.7 x 10' 6 M. The l.P. is then 
determined as [Mg+ 2 ][OH-] 2 = (0.21)(2.7 x IQ- 6 ) 2 ^ 1.5 x 10~ 12 . The K 8P 
is 8.9 x 10~ 12 , so since l.P. < K S1 >, no pptn. occurs. (Ans.) 

Further Regulations of Solubility; Fractional Solubility 

We have seen above that certain ions may be precipitated under given 
circumstances and others held in solution, even though under different 
concentration conditions they too could be precipitated with the added 
reagent. Similar calculations can be used to determine whether or not a 
precipitate once formed will redissolve if factors of concentration and/or 
chemical environment are favorable.* Calculations can show which 
precipitates, if any, in a mixture will dissolve under the given conditions. 
Separations effected in this manner are by a process which can be called 
Jractional solution or fractional solubility. As before, the ratio of the 
solubility of two precipitates involved must be about 10 6 to 1 to get a 
separation clean enough for qualitative analysis. 

Example 8-9. One has a few mg of a mixt. of ZnS and HgS. It is stirred 
with several ml of 2.0 M HC1. Is a separation possible by fract. solub. ? 

Though the H 2 S will be unknown at equil., one might assume as a first approx. 
that one sulfide will dissolve and furnish enough hydrogen sulfide to sat. the 
soln. Then from [H+] 2 [S~ 2 ] = 1.3 x 10~ 21 , the [S~ 2 ] in the 2 M H+ soln. is 
found to be ^ 3.3 x lO" 22 A/, and from the K 81 > of ZnS which is 1 x 10~ 20 the 
equil. [Zn +2 ] is ^ 30 M. This value is so large that the few mg of ZnS in the 
sample will dissolve in the HC1. 

The K SP of HgS is 1,1 x 1Q- 50 and the [Hg+ 2 ] in equil. with the available 
S~ 2 is only = 3.3 x 10~ 29 M. For practical purposes HgS is not going to 

* This assumes the precipitate itself does not change character as some sulfides and 
hydrous oxides do on standing. The K aP values are generally listed only for freshly 
prepared, though well digested, precipitates. 


dissolve at all in 2 M HC1 (or in any other acid soln. for that matter in which 
H 2 S is a by-product) since the K sl , of HgS is so small. 

The Solubility of HgS in Aqua Regia 

After the foregoing calculation one may recall that HgS is separated in 
cation group 2A by fractional solubility, since it is insoluble in HNO 3 , 
while the other members of the subgroup are soluble. The student may 
wonder why in the next step HgS dissolves in aqua regia when before it was 
insoluble in either HC1 or HNO 3 individually. There are two factors 
favoring displacement of equilibrium to give that result: (a) HNO 3 oxidizes 
sulfide to elemental sulfur, a reaction which keeps [S~ 2 ] very low and (b) a 
reasonably stable tetrachloromercury(II) complex forms that keeps [Hg+ 2 ] 
very low. The over-all reaction is 

3HgS + 8H+ + 2NO- + 12C1~ = 3HgCV + 2NO + 4H 2 O + 3S 

It is not correct to say that aqua regia produces oxidants more powerful 
than the original substances. Though NOC1 and C1 2 may be present in 
aqua regia, they are not the whole explanation for the above reaction. 


The student will become aware soon after starting the laboratory work 
that not all precipitates form quickly and settle completely, even after 
heating, prolonged stirring, centrifugation and perhaps some profanity. 
Sulfur and certain metallic sulfides are prone to form many very small 
particles rather than a lesser number of large particles. If the particle 
diameters are in the range 10~ 5 and 10 ~ 7 cm*, the particles are colloidal. 
They will not settle by gravity since they are small enough to be kept in 
motion (Brownian movement) by collisions with solvent molecules and by 
mutual repulsion due to their possessing similar surface charge. They 
are also too small to be filterable by ordinary means. A homogeneous 
appearing suspension like this is called a sol. Particles in this range may 
coagulate and form long strings (micelles) of atoms that trap liberal amounts 
of solvent to give a second type of colloidal suspension called a gel. There 
are a number of familiar gels in analytical chemistry, most of them being 
hydrous oxides, as those of Fe^ 3 , Al f3 , Si TV , and Cr+ 3 . 

Precipitation of Colloids 

From the analyst's viewpoint the most significant feature of colloidal 
particles is their great surface area per unit weight and the ability of that 

* 1 micron, /i = 10 4 cm, 1 millimicron, m/j = 10~ 7 cm, and 1 Angstrom unit, 
A = 10~ 8 cm. These units are used in discussing small dimensions. 


surface to collect (adsorb) ions from solution. Sols adsorb primarily 
ions of one charge or the other, so a given sol will be composed of particles 
whose surface charge is similar. Clustered about this primary layer of 
adsorbed ions is a less tightly held secondary layer of counter ions. 

Sols may be shown in two simple experimental ways to be charged, both 
depending for evidence on the idea that if the surface charge is neutralized, 
coagulation or flocculation of the colloid will take place. The first method 
involves adding a salt to the colloidal suspension to force adsorption of a 
secondary layer, which neutralizes the primary layer. The valences of the 
salt ions of proper charge are in direct proportion to their coagulative 
effectiveness (and one experimental technique for determining the valence 
of complex ions actually uses such an application). If this precipitate is 
then washed with pure water instead of with a dilute salt solution, the 
ionic layer that caused flocculation can be removed and the solid may 
return to its former dispersion by a deflocculation process called peptization. 
The second method of colloid precipitation involves electrolysis of the 
sol ; if it is positively charged, it moves toward the negative electrode and 
coagulation is noted there as the sol's charge is neutralized. This is the 
principle used in precipitating smoke particles in chimneys to prevent air 


Coprecipitation is the term given the process in which a substance that is 
precipitating carries with it other substances from the solution that 
normally would not come down. These coprecipitants and extent of 
contamination by them depend upon the substances involved, their valences, 
the surface area available for adsorption, speed of reaction, temperature, 
time allowed for digestion, etc. There are several mechanisms by which 
Coprecipitation can take place, the most important being that of adsorption 
(described above). Adsorption causes trouble for the analyst in that the 
adsorbed ions may be the ones sought in a later test or they may be such as 
to make confirmatory tests on the main precipitate difficult. Coprecipita- 
tion can be minimized by adding the precipitant slowly and with good 
stirring to a fairly dilute, hot solution. This is followed by a hot digestion 
period and washing of the precipitate with a dilute solution of a volatile 
electrolyte. These precautionary measures favor large, pure crystals 
with minimum surface, and slow tendency toward adsorption and 
peptization. (See problem 25, this chapter.) 

Uses of the Adsorption Phenomenon 

Once the mechanisms of adsorption are understood, conditions can be 
regulated to increase or decrease the phenomenon and it can be put to 


work in the chemist's behalf. One use is illustrated by the adsorption 
complexes (lakes) formed between certain inorganic precipitates and 
organic color reagents (type D, Chapter 12). Another use is in chromato- 
graphic separation methods. In this technique, a solution moves over an 
insoluble medium, on the surface of which solution components concen- 
trate in the order of their ability to be adsorbed. The solution may 
percolate through a column of adsorbant (column chromatographv) or 
move by capillary action over a strip of paper (paper chromatography). 
Adsorbance of colored substances is followed simply by the color banding; 
colorless materials require color reagents or ultra violet light to locate their 
positions. The zones holding adsorbed material are cut out and the 
desired components washed out (eluted) by proper solvents (special experi- 
ment 9). The chromatographic method has recently been applied to 
gases that are identified by thermal conductivity after desorption. 

The analyst may also utilize adsorption to collect or concentrate small 
quantities of precipitates for further testing. For instance, by precipitating 
BaSO 4 on filter paper, one decreases the paper's average pore size to a 
degree that colloidal precipitates (such as Prussian blue from a test for 
Fe+ 3 ) can be removed from suspension by filtration. Detection by spot 
paper techniques is generally of a higher order than detection in solution, 
due partly to the paper's adsorptive effect. 


1. The K SI , of SrCrO 4 is 3.6 x 10 ~ 5 . Arbitrarily select six values of [Sr+ 2 ] 
between 10" 1 and 10~ 4 M and calculate the equilibrium [CrO 4 2 ]. Prepare a 
graph of [Sr 12 ] vs. [CrO 4 2 ]. Explain the hyperbolic shape with reference to 
Sr+ 2 and CrO 4 2 in solution. Explain the significance of the shape of a plot 
of [Sr+ 2 ] vs. l/[CrO 4 ~ 2 ]. What shape curve would one expect to obtain by 
graphing [Pb+ 2 ] vs. [Cl~] 2 in illustrating the PbCl 2 case? In [Pb f2 ] vs. 1/[C1'] 2 ? 
Try these. 

2. A solution is saturated at 20 C by shaking an excess of PbCl 2 in pure H 2 O. 
A 100 ml sample of clear solution is drawn off and evaporated, leaving a residue 
of 990.2 mg of PbCl 2 . Calculate the molar solubility of the salt and the K SP . 
Compare the latter with the table value. 

3. Excess Ca(OH) 2 is shaken with pure H 2 O. A 50.00 ml sample of clear, 
saturated solution is later withdrawn and takes 17.90ml of 0.1400 N HC1 to 
titrate its base strength completely. Calculate the K SJ , of Ca(OH) 2 and compare 
your answer with the text value. How could one have used indicators instead 
of titration to arrive at the K$j, of Ca(OH) 2 ? 

4. Mn+ 2 can be oxidized to MnO^ by trisodium paraperiodate, Na 3 H 2 IO 6> 
in the presence of H 2 SO 4 + H 3 PO 4 . Whereas the color of Mn+ 2 in dilute 


solutions cannot be detected, MnO;*, being a dark purple-red color, can be 
seen at concentrations as low as 0.1 part per million Mn (ppm = mg/liter). 

A sample of MnCO 3 is shaken with pure H 2 O and after the excess has settled, 
a sample of clear, supernatent, saturated solution is withdrawn. To this is 
added cond. H 2 SO 4 , cond. H 3 PO 4 , and Na 3 H 2 IO 6 and the solution is boiled. 
When cool, the color developed due to the oxidation of manganese(II) to 
manganese(VII) is visually compared with a series of known permanganate 
solutions prepared in the same way, and it is decided that the sample matches 
the 0.5 ppm Mn standard. Show that the experiment has given data which 
allows one to calculate the K 8f > of MnCO 3 . (Ans. 8.3 x lO" 11 ) 

5. (a) How many g of (NH 4 ) 2 C 2 O 4 must be added to 1 liter of solution 
containing 0.0050 g of Ca(NO 3 ) 2 to start precipitation of CaC 2 O 4 -H 2 O? If 
0.0500 g of Sr(NO 3 ) 2 were also present, which oxalate would precipitate first? 

(h) What success would one have in attempting fractional precipitation using 
OH" on a solution containing an equimolar mixture of Ca^ 2 and Sr+ 2 ? 

6. MX 2 is a slightly soluble salt. The ionic weight of M 42 is 100 and of X~ 
is 50. The K sl > is 1 x 10~ 12 . (a) Calculate the mg of each ion present in a 
liter of saturated solution. (/>) How many liters would be needed to dissolve 
lOOgof MX 2 ? 

7. Graph the following solubility data for Pbl 2 and calculate the K^ r at each 
temperature. Explain why K K]t is not a constant. 

Temp., C 






Solubility (g Pbl 2 /litcr) 


0.68 % 





8. What is the molar solubility of Ag 2 CrO 4 in pure H 2 O? In 0.02 M AgNO 3 ? 
In 0.02 M K 2 CrO 4 ? Why aren't the last two answers the same since the 
common ion principle is in effect in each and the salt concentrations are the same? 

9. (a) How many mg of Ag f remain unprecipitated when 500 ml of 5 X 10~ 5 
M AgNO 3 is mixed with 500 ml of 5 x 10~ 4 M HC1 ? 

(b) One has 300 mg of AgCl in a sintered glass funnel and washes it with 
300ml of H 2 O. Assuming equilibrium conditions, what weight of precipitate 
remains? Suggest a better wash liquid considering that the residue is to be 
dried later and weighed in a quantitative determination. 

10. A solution is made by dissolving 50 mg of AgNO 3 in a liter of water. 
About how many mg of NaCl are needed to precipitate 20 mg of AgCl from 
this solution? 

11. What is the solubility (g/liter) of Fe(OH) 3 in a solution buffered at/?H 7? 
pW 8? pH 61 If water is flowing through an iron pipe, should one treat the 
water to give it a moderately low or high /?H to maintain a protective coating 
of Fe(OH) 3 ? What is the lowest pH at which one could theoretically precipitate 
Fe(OH) 3 from a 0.1 M Fe+ 3 solution? 

12. A solution is 0.1 A/inH f ,0.01 M in Cd f2 , and 0.0001 A/inCu+ 2 . Which 
sulfide precipitates first when H 2 S is bubbled in ? 


13. A solution contains 2 mg of Mg+ 2 per ml. How many mg of solid 
NH 4 NO 3 must be added to 5 ml of this solution to just prevent Mg(OH) 2 
precipitation when 3 ml of 0.04 M NH 4 OH are later added ? 

14. A solution is 0.120 M in PO 4 3 and 0.055 M in S~ 2 . To this is slowly 
added solid Pb(NO 3 ) 2 . (a) Which lead compound precipitates first? (/>) When 
[PCV 3 ] becomes 10~ 4 M, what is [S~ 2 ]? (r) When [S~ 2 ] is 1Q- 10 M, what is 
[P0 4 - 3 ]? 

15. If to 1 liter of 0.010 M NaBr is added 10.000 g of solid AgCl, what is the 
ratio [Br~]/[Cl~] at equilibrium (assuming no volume changes)? Does the 
magnitude of the ratio suggest that one could successfully separate these silver 
halides by fractional precipitation? 

16. Show that if one keeps 10ml of solution, which is 0.3 M in H + and 
0.01 M in Cu+ 2 , saturated with H 2 S, a quantitative precipitation of CuS is 
obtained. Show by calculation that ZnS is not precipitated if the same solution 
also contained a [Zn+ 2 ] of 0.01 M. Calculate the final total [H+] when the 
[Cu+ 2 ] has been reduced to 10~ 10 M. 

17. Determine what [H f ] is needed from HC1 to begin solution of CuS. Of 
ZnS. Does this calculation show that one might effect a separation between 
the two sulfide precipitates by [H+] control, say with a buffer solution or by 
using a weak acid? (Assume saturated H 2 S solutions.) 

18. Can one dissolve an appreciable quantity of FeS in 1 M HAc? What is 
the solubility of FeS in a buffer solution made by dissolving 8.20 g of NaAc 
in a liter of I M HAc? (Same assumption as 17.) 

19. A solution is 0.4 M in Ba 42 and 0.15 M in Sr+ 2 . If solid K J SO 4 is 
slowly added, and assumed to dissolve and mix immediately (a) At what 
[SO^ 2 ] does BaSO 4 start to precipitate? (b) What is the ion product [ST H2 ][SO 4 2 ] 
when condition (a) obtains? (c) What is the [Sr+ 2 ] when [Ba+ 2 ] is 10 M? 
(d) What is the [Ba* 2 ] when two-thirds of the Sr K2 has precipitated? (Ans. 
() 3.75 x 10~ 9 A/.) 

20. Ten ml of solution contains 17 mg of Ag + , 20 mg of Cu+ 2 , and is 0.20 M 
in H + . It is kept saturated with H 2 S and both metallic sulfides for practical 
purposes are quantitatively precipitated. Not neglecting the H 1 formed in 
the reactions, calculate the equilibrium concentrations of H+, Cu+ 2 , Ag 1 and 
S" 2 . (Ans. [H+] = 0.279 M.) 

21. From K. t of HNO 2 and K Nl > of AgNO 2 , show that K for AgNO 2 + H+ 
= Ag+ + HNO 2 has a value of about 0.27. Would you expect to dissolve 
much AgNO 2 at, for example, />H 5 ? Explain. 

22. (a) One has a very fine water suspension of BaSO 4 from which he \\ishc- 
to recover the solid. How would the addition of a small amount of agar gel 
followed by centrifuging help? How could one increase the size of the BaSO 4 
particles so they will settle faster of their own accord? 

(h) Iron pyrite (FeS 2 + SiO 2 ) is analyzed by dissolving FeS 2 with a mixture 
of HNO 3 + Br 2 , which oxidizes sulfide sulfur to sulfate. This is then preci- 
pitated as BaSO 4 . After the precipitate is ignited (to dry it before weighing) it 
invariably comes out reddish colored and one obtains results that are too high 
for sulfur unless a correction in technique is made near the beginning of the 


process. Explain what has happened and how to circumvent the difficulties so 
as to obtain correct analytical results. 

23. If AgCl is precipitated from solution containing excess AgNO 3 and dried, 
weighed, and analyzed for silver, it is found to contain a little more than the 
theoretical amount of metal. With the aid of a suitable diagram, give an 
explanation for the experimental result. 

24. (a) Express the colloid particle diameter range in mm, //, m// and A units. 
(b) The density of Pt is 21 .45 g/cc. Calculate the cm 2 area of a g cube of Pt. 

If this cube were reduced in size to uniform cubes, 100 m/* on a side, calculate the 
new surface area. If this powder were used as a catalyst for a reaction that 
took place at the catalyst's surface, by what factor would you estimate the process 
may be speeded up by sub-division of the solid? 

25. (Library) (a) Two mechanisms of coprecipitation are occlusion and 
mixed crystal formation. With the aid of a textbook in quantitative analysis, 
describe these processes and point out how they apply to qualitative testing. 

(b) NH 4 NO 3 is frequently used as an electrolyte in aqueous washes to prevent 
peptization. Why is it superior to NaCl, for example, if the precipitate is to 
be later heated and weighed quantitatively? 

26. (Library) Find reference to these instruments mentioned in the chapter: 
turbidimeter, nephelometer, and ultramicroscope. Give a schematic drawing 
of each, label, and briefly explain how each works. Mentally note how they 
are related to K sr determinations. The ultra-microscope method is reputed 
to be the least reliable of the three. Can you suggest any reasons for its 

27. A solution is 10~ 5 M in Ag* and another is 0.152 M in CrO^ 2 . Jackson 
Slipshod wants to find if upon mixing equal volumes, silver chromate will 
precipitate (K SP = 1.9 x 10~ 12 ). He reasons first that the solutions mutually 
dilute each other, so momentarily [Ag+] = 5 x 10 6 M and [CrO 4 2 ] = 0.076 M. 
Then to get the ion product, the silver concentration is doubled and squared 
and multiplied by the chromate concentration: I. P. (10 x 10 H ) 2 (7.6 x 10 2 ) 
= 7.6 X 10 12 . Since LP. > K sr , Ag 2 CrO 4 should comedown. On actually 
mixing the solutions nothing happens and Jackson concludes that the table K SJ > 
value was in error. What alternate explanation can you find? 


1. R. K. McAlpine, J. Chem. Educ., 23, 28 (1946). (Precipitation sensitivity) 

2. K. B. Morns, J. Chem. Educ., 24, 236 (1947). (K KI . and the Ag halides) 

3. E. A. Mauser, /. Chem. Editc., 25, 367 (1948). (Colloid nomenclature) 

4. J. A. Bishop, J. Chem. Educ. 31, 574 (1954). (K s /> graphs) 

5. J. E. Land, /. Chen. Educ., 34, 38 (1957). (Nephelometry and formulas of 

6. L. Gordon, Anal. Chem., 24, 459 (1952). (Precipitations) 

7. J. H. Hildebrand and G. J. Rotariu, Anal. Chem., 24, 770 (1952). (Solubility 

There are many colloid texts and reference books on special topics in colloid science. 



Discussion in Chapter 4 showed that Werner salts are like the simpler 
salts in the sense that they are ionic and dissociate in aqueous solution. 
It was left until now, however, to point out that the complex ions them- 
selves are capable of undergoing a second, much weaker ionization and 
that* the chemical equilibrium principle is operative and applicable. 
Factors in this ionization are the ion's stability as dictated by the bond 
strengths, temperature, solvent, concentration, and the presence of other 
dissolved substances. By considering only variation in ion identity and 
concentrations, we are able to learn some fundamentals of these systems in 
a quantitative way. Equilibrium applied here is another special case of 
the mass action law. 

Dissociation of Complex Ions 

The first dissociation of a Werner compound is considered to be com- 
plete in dilute solution : 

Cu(NH 3 ) 4 S0 4 = Cu(NH 3 )+ 2 + SO 4 2 

As with other strong electrolytes, the equilibrium principle is not adaptable 
to this. The tetrammine copper(Il) ion will, however, then exhibit a 
second dissociation, yielding some ammonia and cupric ion in solution: 

Cu(NH 3 )+ 2 = Cu^ 2 + 4NH 3 

This dissociation takes place to a small degree and since it varies in a 



predictable inverse ratio to the concentration of the complex, one may 
further develop and use the mass action concepts to aid understanding of 
such examples. The equilibrium constant is called the instability con- 
stant, K in89 and is a measure of the ion's ability to dissociate: 

K tns = [Cu+*][NH 3 ]V[Cu(NH 3 )+ 2 ] 

It is evident that the concentration in solution of the coordinated groups 
for all these complexes will be an important factor in shifting equilibrium 
because of the many cases where 4 and 6 appear as powers. 

Determination of K tn8 Values 

In Chapter 4 a number of experimental techniques were explained which 
have been useful in determining the formulas for complexes. After 
formulas have been found it is of interest to analytical chemists to evaluate 
the instability constants of the complex ions to see how tightly bound the 
complexes are and of what use one can make of this knowledge. Zinc 
and aluminum ions can be separated, for instance, by adding aqueous 
ammonia; zinc forms the soluble, stable complex Zn(NH 3 )^ 2 , whereas 
aluminum precipitates quantitatively as A1(OH) :} (hydrated). Another 
application in group 3 analysis is the complexing of ferric ion with phos- 
phoric acid, so the thiocyanale-acetone test for cobalt will not be obscured 
by any Fe(SCN) j 2 color, etc. If instability constants are known for com- 
plexes, computations can be carried out which make these processes more 

Several simple methods have been used to find equilibrium constants 
for the Werner ions, and two arc briefly dealt with below. 

1. K ins from Electrochemical Cells. Proper cell setups can relate 
measured voltage (which is dependent upon ion concentration) to the 
instability constant of the system studied. This will be illustrated in 
Chapter 11. 

2. K 1tlii /rom Solubility Data. One can sometimes relate equilibrium 
constants that are known to an instability constant that is unknown and 
evaluate the latter. K sr data have been helpful in this respect. 

Example 9-1. A student finds that at room temp, the solub. of solid AgCl 
in 0.022 M NH 4 OH is 10~ 3 moles/liter. Show that this is enough data for a 
determination of the instability constant of the diammine silver complex. 

Two equilibria are involved: 

AgCl = Ag^ + C1-; K 81 , = [Ag+][CI-] = 2.8 x 10- (9-1) 

Ag(NH 3 )+ = Ag+ + 2NH 3 ; K m = [Ag^NH^/tAgfNH^ ] = ? (9-2) 

The reaction is 

AgCl + 2NH 3 = Ag(NH 3 ) 2 + + Cl~ 


At equil., [Cl~] = 10~ 3 M and [Ag(NH 3 )^] = 10~ 3 M if one assumes that all 
10~ 3 moles of Ag f are complexed. This is a valid approx. since the complexes 
used in anal. chem. are usually quite stable. Then [NH 3 ] = 0.022 - (2)(1Q- 3 ) 
= 0.020, since 2 NH 3 's are used to complex one Ag+. With these figures one 
substitutes first in equation 9-1 to find the equil. [Ag+] and then in equation 
9-2 to get the instability constant: 

[Ag+] = 2.8 x 10- 10 /10- 3 = 2.8 x 10~ 7 M 

In this homogeneous soln. there is only one silver ion concn. and it must satisfy 
both equations. The other concns. being known, one writes 

K ins = (2.8 x 10 7 )(2 X 10~ 2 ) 2 /10- 3 = 11.2 X 10~ 8 (Ans.) 

Example 9-2. A certain soln. is 10~ 4 M in total Cd+ 2 and 0.90 M in HC1, 
and when this mixt. is satd. with H 2 S, CdS just begins to ppt. Show that a 
discrepancy exists between these actual conditions of pptn. and the theoretical 
conditions. Assuming this is due to the presence of a complex, CdClJ" 2 , calc. 
a value for its instability constant. 

From equation 7-23, we get, for satd. H 2 S solns., [H f ] 2 [S~ 2 ] = 1.3 x 10~ 21 , 
and the [S" 2 ] in equil. with 0.90 M H+ is calcd. to be 1.61 x 10~ 21 M. (As in 
the preceding chapters on equil. it is assumed that the HC1 is completely dissoc.) 
From the K sr of CdS, [Cd 1 2 ][S~ 2 ] = 6.0 x 10~ 27 and now knowing [S~ 2 ], one 
finds that the theoretical amount of Cd * 2 needed to just start CdS pptn. is 
3.7 x 10~ 6 M. According to the experimental evidence, however, I x 10~ 4 M 
Cd +2 is actually needed. 

If one postulates the existence of a tetrachlorocadmium(II) complex (since 
Cd+ 2 is known in other instances to coordinate four groups) then this might 
account for the difference between the theoretical and actual cadmium concns. 
needed for CdS pptn. : 

CdClr 2 = Cd- 2 + 4C1- 

The discrepancy between the two [Cd 12 ] values must equal the tetrachloro 
complex concn. since that amount of cadmium is apparently tied up as metal 
ions unavailable for pptn. The equil. concns. are: 

] = 10 4 - 3.7 x 10- 6 ^ 9.6 x 10 5 M 

[C1-] ^ 0.90 M (since little CI~ is used in the small quantity 
of complex formed) 

[Cd+ 2 ] = 3.7 x 10 6 M 
The instability constant is then deducible: 

K in8 = [Cd+ 2 ][Q-] 4 /[CdCl 4 " 2 ] = 2.5 x 10~ 2 (Ans.} 

The relatively large constant shows this complex is not as stable as 
many that are known. Moderate dilution for example would upset the 
equilibrium enough to allow considerable cadmium to precipitate as the 
sulfide. One Cu |2 -Cd+ 2 separation is based on this idea; excess NaCl is 


shaken with the mixture, then H 2 S added. Copper coordinates Cl~ 
quite weakly and the K SP of CuS is very small, so CuS precipitates, but 
because CdCl^ 2 , and possibly other chloro complexes, are fairly stable 
(particularly in the presence of excess NaCl which forces the equilibrium 
toward the complex), CdS does not come down simultaneously. If the 
mixture is centrifuged and the centrate diluted, the equilibrium shifts to 
give more Cd+ 2 in solution and enough H 2 S remains dissolved to now 
cause yellow CdS to appear a characteristic color that is otherwise 
obscured by black CuS. 

In calculations, one runs into difficulty if several different complexes 
form as CdCl+ and CdClg . In such cases one generally does not have 
enough data to give more than an approximate solution to the problem. 

3. K ina from Partition Data. As mentioned in the section on physical 
equilibrium in Chapter 6, a solute distributes itself between two immiscible 
solvents when shaken with a mixture of them, in the ratio of its solubility 
in each. This ratio is called the partition, or distribution coefficient, 
K D , and is established by measuring the concentration of solute in each 
layer by some simple means like color density matching or titration. If 
this equilibrium constant is previously experimentally determined and the 
concentration of solute in one solvent is known, then the concentration of 
the solute in the other solvent is simply obtained by arithmetic 

K D = Concn. in solvent A/concn. in solvent B (9-3) 

This establishes a known value without the need of direct measurement, 
and if that solute is involved in a reaction with complex ions and means 
are available for finding the concentrations of other species in that solvent, 
one may be able to not only deduce what reaction is taking place and 
what complex is present, but also the instability of the complex. The 
general method has been useful in many investigations and qualitative 
separations such as those illustrated in tests 22-21, 22-34, etc. 

Example 9-3. K D for I 2 in CC1 4 and H 2 O at 20 has been experimentally 
found to have a value of 80: 

KD = [Ic- 4 )]/[W = 80 (9 ~ 4 > 

A soln. contg. a known amount of KI in water is shaken with a carbon tetra- 
chloride soln. 0.060 M in I 2 . Equil. of I 2 between the solvents is established 
after a time and one notices that the water layer has assumed an orange color. 
It is postulated that reaction takes place between I 2 and I~, giving I 3 " to account 
for increased aqueous I 2 solub. and the color. In a certain expt., 0.1000 M I" 
soln. is used and after being shaken with the iodine-carbon tetrachloride soln., 
a titration with standard sodium thiosulfate (see test 22-39) of the water layer 
shows that the concn. of Ijf plus I 2 (both of which react with S 2 Ojf 2 quantitatively) 


is 0.0028 A/. In a later, similar expt. to gather data at different concns., when 
0.080 M I 2 in CC1 4 was shaken with 0.1200 M KI aq. soln., the equil. water layer 
showed upon thiosulfate titration a concn. of I 3 ~ plus I 2 of 0.0042 M. From 
these data, easily gathered in the lab. in a few hours, one is able to prove that 
the postulated reaction is the correct one and he is also able to evaluate the 
instability constant for \%. 

Using the first set of data one assumes in the beginning that the CC1 4 soln. 
is sufficiently coned, in I 2 so that the quantity dissolving in the water will not 
alter the 0.06 M concn. much. This is a valid approx. since K D was a compara- 
tively large number. 

Pica*)] Bcci 4 )]/80 = 0.00075 M 

Since titration gave [1^] 4~ [I 2 (aq)l 0.0028 A/, the triiodide concn. can be 
found to be 0.0028 0.00075 ^ 0.002 M. (Notice that without the pre- 
determined distribution constant this would not be possible, since thiosulfate 
titration does not differentiate between I 2 and 1^.) The original [I"*] in the 
water layer is known (I"" is not sol. in CC1 4 ), since the original soln. was made 
0.1000 M in I~ and from this and the triiodide value deduced above, one can 
find the cquil. [I~]. 

Assuming the reaction in water to be 

one can determine the concn. of iodide remaining. Iodide was originally 
0.1000 A/, but if it reacted with some I 2 to form 1^, then at equil. [I ] + [1^] 
= 0. 1000 M. Since [1^] ^ 0.002 A/, [I ] ^ 0.098 M. With the equil. concns. 
of all species known one can determine the instability constant of I a , , because 
its dissoc. is the opposite of its synthesis proposed in equation 9-5: 

= (0.00075)(0.098)/0.002 = 0.0368 (Ans.) 

Using the second set of data mentioned above, one obtains almost the 
same for K tns , and several other verifying experiments give the same 
closely checking results. One concludes from this that equation 9-5 
correctly states the reaction and equation 9-6 is the statement for the 
complex ion dissociation. If one assumes on the other hand that the 
reaction was I 2 + 2I~ = Ij 2 , it is easy to show that K ins is not a constant 
and the proposal is false. 

Complex Ions and the Cu t2 -Cd+ 2 Separation 

Further application of the type of calculations in the foregoing para- 
graphs is found in the copper-cadmium separations of the group 2 pro- 
cedure. By one method, CN~ in excess is added to the ammoniacal 


solution of the metallic ions, converting Cu(NH 3 )+ 2 and Cd(NH 3 )+ 2 to 
Cu(CN) 3 2 and Cd(CN)j 2 . When H 2 S is introduced into this mixture, 
only CdS precipitates, the instability constant of CdfCN)^ 2 alone being 
large enough to furnish sufficient metallic ion in solution for its sulfide 
solubility product constant to be exceeded. The K 8l > of Cu 2 S is very 
small itself (see Table 22-1), surprisingly enough about ID" 22 times smaller 
than that of CdS, but because the K ins of the tricyano copper(I) ion is 
also very small, Cu 2 S is not precipitated, and thus separation of the two 
metallic ions is made. Example 9-4 shows how this is possible. 

Example 9-4. A basic soln. is 10~ 2 M in both Cd(CN)^ 2 and Cu(CNV 2 
and is also 0.1 M in CN~. H 2 S is bubbled in until the concn. of S~ 2 is 0.1 M. 
Show that CdS, but not Cu 2 S, ppts., affording a means for the separation of 
metal ions. 

Solving first for the cadmium concn. available from dissoc. of the complex: 

X ins = 1.4 x 10- 19 = [Cd+ 2 ][CN-] 4 /[Cd(CN)4T 2 ] = [Cd+ 2 ](0.1) 4 /10- 2 

[Cd f2 ] = 1.4 x 10~ 17 A/ 

The ion product [Cd+ 2 ][S- 2 ] = (1.4 x 10- 17 )(0.1) = 1.4 x 10 18 . Since this is 
larger than the K SP of CdS, which is 6 x 10 27 , CdS ppts. (Ans.) 

The cuprous complex is more stable and less copper ion is available than 

K, ns = 5 X 10- 28 = [Cu+][CN-] 3 /[Cu(CN)3- 2 ] = [Cu+](0.1) 3 /10- 2 
from which 

[Cu+] = 5 x 10~ 27 M 

The l.P. of cuprous sulfide = (5 x 10~ 27 ) 2 (0.1) = 2.5 x 10~ 54 , which is smaller 
than the corresponding K AS/ , of 1.2 x 10~ 49 , so Cu 2 S cannot ppt. (Ans.) 

This and other ways of detecting these metals in the presence of each 
other are given in Chapter 1 6. 

More Analytical Applications of Werner Ions 

Separations of ions by complex formation and stabilization of certain 
oxidation states in solution are common applications of Werner ion 
chemistry. Two examples follow. 

(1) From a mixture of chloride and iodide, only Agl is precipitated by 
addition of a solution containing a definite excess of NH 3 plus Ag(NH 3 ). 
The K SP of silver iodide being about 10~ 6 times smaller than the K HP of 
silver chloride, it alone is exceeded and the single halide precipitates with 
the controlled Ag+ concentration the complex releases to the solution. 


(2) Cyanide may be determined quantitatively by titration with a 
standard silver nitrate solution. As long as CN" is present, the reaction 
Ag + + 2CN~ = Ag(CN)^ takes place. The silver complex is soluble 
and colorless. At the end point, however, the first excess Ag+ gives a 
white turbidity due to the reaction, 

Ag^ + Ag(CN) 2 - = 2AgCN 

Apropos of the first of these applications, another calculation should 
be illustrated. 

Example 9-5. (a) If to a soln. 0.200 M in uncomplexed NH 3 and 0.020 M 
in Ag(NH 3 )^ is added an equal vol. of 0.002 M NaCl, will AgCl ppt. ? 

From the data given one can calc. the ion product of AgCl and compare it 
with the K SP of AgCl as the condition for pptn. Each soln. dilutes the other: 

K ms = [Ag+][NH 3 ] 2 /[Ag(NH 3 )+] = 5.9 x 10' 8 
[Ag+] = (5.9 x 10- 8 )(0.010)/(0.100) 2 = 5.9 x 10~ 8 A/ 

Then, since [Cl~] = 10~ 3 A/, I. P. = 5.9 x 10~ n , which is smaller than the K SP 
2.8 x 10- 10 , so AgCl will not ppt. (Ans.) 

(b) If instead of NaCl, 0.002 M Nal is added under the same conditions, can 
Agl form ? The I.P. will be the same, 5.9 x 10 -", since that part of the problem 
has not changed. The K SP of Agl, however, is 8.5 X 10~ 17 and now I.P. > K s/ > 
so Agl ppts. (Ans.) 

Something of a corollary to this problem is the following. 

Example 9-6. Calc. the wt of AgBr that one can dissolve in 1 liter of 2 M 
NH 3 . 

Two ways will be illustrated in solving this one. 

(Method 1.) Since the K SP of AgBr is quite small, 5 x 10~ 13 , one may assume 
that little will dissolve even though the ammonia complex can form and its K ins 
is 5.9 X 10~ 8 . Therefore the equil. [NH 3 ] ^ 2 M. The reaction being 

AgBr + 2NH 3 = AgCNH.,)^ + Br~ (9-8) 

it is seen that [Br~] ^ [Ag(NH 3 )^] because the value of K tns shows that the 
complex has reasonable stability and does not dissoc. to yield much free Ag+ in 
soln. If these figures are substituted in the K SP and K ins expressions, the results 

#S P = 5 x 10- 13 = [Ag+][Br-] = [Ag+][Ag(NH 3 )+] 

K tns = 5.9 x 10- 8 = [Ag+](2) 2 /[Ag(NH 3 ) 2 +] 
Solving each for [Ag+], the second equation gives 

[Ag+] = [Ag(NH 3 ) 2 +](1.5 x 10~ 8 ) 
and the first gives 

[Ag+] = 5 x 10- 13 /[Ag(NH 3 ) 2 +] 


Because the soln. is homogeneous in [Ag+], these two equations are equal to 
each other, or [Ag(NH 3 ) 2 l ](1.5 x lO' 8 ) = 5 x 10- 13 /[Ag(NH 3 ) 2 + ]. From this, 
one finds that the concn. of the complex is about 5.8 x 10~ 3 M. Since most 
of the sol. silver is in this form, this is approx. the molar solub. of AgBr in 
2 M NH 3 . The g dissolved are obtained by multiplying by 188, the mol. wt, 
and the salt = 1.1 g AgBr. (Ans.) 
(Method 2.) If one writes the equil. constant for equation 9-8 

K = [Ag(NH 3 ) 2 f ][Br-]/[NH 3 ] 2 (9-9) 

he finds K can be evaluated in terms of the equil. constants discussed above: 

K 8P lK in , = [Ag^][Br-]/([Ag+][NH 3 ] 2 /[Ag(NH 3 ) 2 + ]) 
and cancelling [Ag + ] and rearranging gives equation 9-9. 

K = K 8V lK m , = 5 x 10~ 13 /5.9 x 10" 8 = 8.5 x 10~ 6 
By the same reasoning used above and having [NH 3 ] given as 2 A/, 

8.5 x 10- 6 ^[Ag(NH 3 )^] 2 /(2) 2 
from which the same answer as above is obtained. 


1. Jackson P. Slipshod announces a new separation for Cu+ 2 and Cd+ 2 . 
"Make the solution 0.10M with respect to uncomplexed NH 3 and 0.01 M 
with respect to Cu(NH 3 ) 4 1 2 and Cd(NH 3 ) 4 ' 2 . Sprinkle in enough Na 2 S to 
make the [S~ 2 ] 10 3 M momentarily. Because the K ins of the Cd complex is 
larger than that of the Cu complex, only CdS precipitates under these conditions." 
Check the basis for the Slipshod approach. 

2. A solution is 0.1 M in [AgCNH^ ] and additional NH 4 OH gives the solution 
a pH of 10.0. Calculate the [Ag f ], 

3. Find the approximate concentrations of all species in a solution 0.1 M 
in Ag(NH 3 ) 2 NO 3 . (Ans. 0.1 M NO^, 1.1 X 10~ 3 M Ag+, 2.3 x 10~ 3 M 
NH 3 , 0.1 A/Ag(NH 3 ) 2 f .) 

4. (a) What weight of AgBr can be dissolved in 100 ml of 6 M NH 4 OH? 
(b) What must the concentration of NH 4 OH be if 1 g of AgBr is to dissolve 

in 100 ml. ? (What is M of coned. NH 4 OH in the lab. ?) 

5. If to 1 liter of the solution in problem 3 is added 1 mg of solid NaBr, will 
AgBr precipitate? Demonstrate with calculations. 

6. How many ml of 0.05 M Ag + are needed to just complex 5 moles of CN~ 


7. A student has a residue of AgCl weighing 600 mg on filter paper in a funnel, 


and washes it with 50 ml of 10~ 3 M NH 4 OH solution. How many mg of 
AgCl are left after this treatment, assuming equilibrium was established in the 

8. Predict whether or not: 

(a) Zn(CO 3 ) will dissolve in 1 M NH 4 OH. 

(b) Hg(IO 3 ) 2 will dissolve in 1 M Br~. 

9. A solution contains 0.15 moles of AgCl dissolved in 1 liter of 1 A/NH 4 OH. 
Tests with a sensitive reagent show the [Ag 4 ] in the soln. is about 10~ 3 ppm 
(mg/liter). Show that this data leads to an approximation of the instability 
constant for [Ag(NH 3 )j-]. (Ans. 3.0 x 10~ 8 ) 

10. In Example 9-3, prove that the reaction equation is not 2I~ + I 2 = I^ 2 . 

11. Fifty ml of 0.1 M [Ag(NH 3 ) 2 ] solution gives a white precipitate with a 
certain amount of NaCl. The same vol. of 0.1 M [Ag(CN)^~] does not give a 
reaction with the same quantity of salt. Explain. 

12. Could one devise an analytical separation of AgCl and AgBr on the basis 
of the former being more soluble in 2 M NH 4 OH? Explain quantitatively. 

13. Calculate the values for the equilibrium constants of 

(a) A1(OH) 3 + 6F- = [AlF^ 3 ] + 3 OH~ 

(b) Hg(I0 3 ) 2 + 4SCN- = [Hg(SCN)4 2 ] + 2IO 3 

From the magnitudes of K, make qualitative statements concerning the 
solubility of these solids in the complexing agents. See Table 15-4 and the 
section on Al in Chapter 17. 

14. Silver bromide has been precipitated in several flasks and a film of it still 
remains after soap and water washing. Rank NH 4 OH, CN", and S 2 O3~ 2 in 
order of decreasing effectiveness as glass cleaning solns. Explain. 

15. By two methods, determine how many liters of 1 M NH 4 OH are needed 
to dissolve 1.0 g of CuS by formation of tetraamine copper(II) ion. 

16. In electroplating one wants the metal ion in low concentration so that 
plating will be slow and a hard plate will be produced, yet the solution must be 
heavily populated with ions so it conducts well. Explain why copper and 
cadmium are frequently plated from high pH cyanide baths. What would 
happen at low pH ? 

17. (a) One suggestion for washing off streets contaminated with radio 
cobalt ions is the use of dilute ammonium hydroxide. Explain. 

(b) One "dip-type" cleaner used to remove Ag 2 S from tarnished silver 
contained NaOH + NaCN, but was taken off the market because of its toxicity. 
What chemistry was behind its cleaning action ? 

18. You are asked to make some preliminary calculations on a reclamation 
process for washing unreacted AgCl from a special film using 1 M sodium 
thiosulfate. If each foot of this 35 mm film contains an average of 10.0 mg of 
AgCl, does it appear feasible that a liter of thiosulfate will treat several feet of 
film and give a solution concentrated enough in silver to make metal recovery 

19. In a student experiment it is found that 1.00 ml of 3.00 M NH 4 OH just 
dissolves 230 mg of Ag 2 CrO 4 . Given the K sl > of that salt (a) find the equilibrium 


[NH 3 ], [Ag(NH 3 #],and [CrOj 2 ] and (b) from the data, find K ins for the complex. 
(Ans. (a) 0.228, 1.386, 0.693). 

20. Solve the problem of Example 9-3 using the second set of data given 
there and compare answers. 


1. S. Z. Lewin and R. Seider- Wagner, J. Chem. Educ., 30, 445 (1953). (Fe (3 -SCN" 

2. C. R. Johnson, /. Chem. Educ. 31, 205 (1954). (Ag salts in NH 4 OH) 

3. P. W. West and C.G. DeVries, /to"/. C/ze/w.,23,334(1951). (Co t2 -SCN~ reaction) 
(Also see list at the end of Chapter 4) 





The term hydrolysis is loosely applied to any reaction in which water is 
split or is otherwise a reactant. As used here, hydrolysis will be taken 
to mean the reaction between water and salts to produce weak acids and/ or 
weak bases. Thus reactions of the type shown in equation 10-1 will not 
be considered, but those illustrated by equation 10-2 will be 

4H 2 + 3Fe = Fe 3 O 4 + 4H 2 (10-1) 

H 2 + NOj = HN0 2 + OH- (10-2) 

The hydrolysis process is reversible and usually only a few per cent 
complete unless the products formed are very weak electrolytes. Again 
the mass action principle is operative and the methods of problem solving 
given in the previous four chapters will be used again. The equilibrium 
constant for the hydrolysis reactions is another limiting case of the general 
equilibrium constant and will be called the hydrolysis constant and 
abbreviated, K n . 

Determination of K H 

Hydrolysis constants are found experimentally by means that have 
been described in earlier chapters on equilibria. These procedures include 
(a) measurement of solution />H, which will be shown to bear a simple 
relationship to K H (b) conductivity data, which includes evaluation of the 
individual contribution to total conductivity by each species present to 
find the equilibrium concentrations (this is not always easy to do), and (c) 



partition measurements, from which one may again get equilibrium informa- 
tion. Some of these methods will be explained or illustrated with 

Hydrolysis of Neutral Salts of Strong Acids and Strong Bases 

Neutral* alkali and alkaline earth salts of HC1, HBr, HI, HC1O 4 , 
HNO 3 , and H 2 SO 4 are the main types considered under this heading. 
Possible reaction, taking KNO 3 as a typical salt, is 

K+ + NO- + H 2 = K+ + OH- + H+ + NO^- (10-3) 

But since KOH is a strong base the ions composing it are not prone to 
combine in solution and the same can be said about the strong acid HNO 3 . 
Therefore the dissociation of water is unaffected, since the salt ions do not 
react with either H+ or OH~. According to the discussion on water in 
Chapter 6, this means that the/>H of such solutions will be 7 since H + and 
OH" are produced in equal numbers and 

[Ht][OH-] = 1 x 10- 14 = K w (10-4) 

Neutral (strong) salts thus do not hydrolyze since they cannot shift the 
water equilibrium by a demand for hydrogen or hydroxyl ions. 

Hydrolysis of Salts of M onvalent Weak Bases and Strong Acids 

The only common weak base for purposes here is ammonium hydroxide 
and a typical example of a weak base-strong acid salt is NH 4 NO 3 . This 
gives ammonium and nitrate ions in solution, but only NHJ is capable 
of reaction with water (hydrolysis) since it is a derivative of a weak 
electrolyte. The reactions are: 

NH+ + NO" + H 2 = NH 4 OH + H+ + NO 3 ~ (10-5) 


NH+ + H 2 O = NH 4 OH + H+f (10-6) 

The equilibrium expression in equation 10-6 will not include water, as its 
concentration is regarded as unchanging, so the hydrolysis constant 

K u = [NH 4 OH][H+]/[NH+] (10-7) 

* A neutral salt implies here a strong acid-strong base derivative giving a solution pH 
of 7. Acid salts like NaHSO 4 give acidic aqueous solutions. 
t This can also be written 

NH+ + H 2 = NH 3 + H 3 0+ 
Acid 1 + base 2 = Base 1 + acid 2 (10-10) 

to show the role of H 2 O in functioning as a Bronsted base. This representation of 
hydrolysis equations will not be used for the reasons given in the footnote on p. 86 
though it is to be understood that the bases NH 3 and H 2 O compete for the proton. 


K H is related to K w and K B , since by multiplying both numerator and 
denominator by [OH~] (a method used before in example 8-7), one 

K H = [NH 4 OH] [H+] [OH-]/[NH+] [OR-] (10-8) 

which can be regrouped : 

K H = [H+] [OH-] - [NH 4 OH]/[NH+] [OH~] ( 1 0-9) 

The first part is AT, r , the water constant, and the second is 1/A^, the 
reciprocal of the ionization constant of the weak base. Substituting these 
known quantities into equation 10-7, one obtains a general equation for 
this type salt. 

K H = K W \K B (10-11) 

From equation 10-6 it can be seen that since one ammonium hydroxide is 
formed for each proton [NH 4 OH] = [H 1 ]. Further, the salt in solution 
is 100% ionic and if each ion directly derived from it is considered to be 
always capable of independent action, then for monovalent ions [salt] 
= [each ion]. If one assumes that hydrolysis is only a few per cent 
complete, as it is with usual dilutions of most salts, then the- original salt 
concentration will not change much due to reaction of the cation with 
water, so 

K n ^ [H+] 2 /[NHJ], or [H+] ^ (* H [NH+])1 

If c is the molar concentration of the cation, a general formula can be 
developed for finding the hydrogen ion concentration of a solution of a 
salt whose cation is derived from a weak base : 

[H+] ^ (K n -c)l z* [(K ir /K*)c]i (10-12) 

The [H+] varies directly with the square root of the concentration of salt 
but inversely with the square root of the ionization constant of the weak 
base. The simple relationships between /?H, c, and K H immediately 
suggest that constants of this type are determinable by merely measuring 
solution /?H and vice versa. 

Example 10-1. 50 ml of 0.2 M HNCX, are titrated with 50 ml of 0.2 M 
NH 4 OH. Calculate the [H+] and /?H at the equivalence point. What indicator 
would one use? What is the % hydrol. of NH^ in the final soln.? 

At the end point one has a soln. 0.1 M in NH^, since NH 4 NO 3 is the product 
formed and the reacting solns. dil. each other. From (10-12): 

[H+]^ [(I x 10~ 14 /1.8 x 10- 5 )(0.1)]*^7.5 x 10~ 6 M (Ans.) 
From the definition of pH in Chapter 6 

pH = - log [H + ] = log 10 6 - log 7.5 ^ 5.1 (Ans.) 


From Appendix A 17, one notes that bromcresol green and methyl red have 
ranges which include pH 5. 1 . (Ans.) 

The degree of hydrol. is that fract. of the salt ion that has reacted with water. 
The percent of hydrol. is 100 times that. The amount of NH^ hydrol. is equal 
to the amount of H+ formed, therefore 

(7.5 x 10- 6 /0.1)(100%) = 7.5 x 10- 3 % hydrol. (Ans.) 

Hydrolysis of Salts of Monovalent Weak Acids and Strong Bases 

Salts of this type are formed by reaction of alkali and alkaline earth 
bases with the weak acids listed in Appendix A20. The remarks and 
procedures in the section just previous will apply again, with a few 

KAc, potassium acetate, is a weak acid-strong base salt. K+ does not 
react with H 2 O, as shown in equation 10-3. The Ac~, however, does 
hydrolyze because the product, HAc, is a weak acid. The reactions are: 

K+ + Ac- + H 2 O = K+ + HAc + OH~ 

Ac- + H 2 O = HAc + OH- (10-13) 

and the hydrolysis constant is 

K H = [HAc][OH~]/[Ac-] (10-14) 

Again K H is related to equilibrium constants of the substances present, 
in this instance to K w and K A . By multiplying the right side of equation 
10-14 by the factor [H+]/[H+] and separating the result into two parts 
one gets : 

K n = [H+][OH-][HAc]/[H+][Ac-] 

The first product is recognized as K w , the water constant, and the second 
as \IK A , the reciprocal of the weak acid's ionization constant. Sub- 
stitution gives 

K u = K W IK A (10-15) 

From equation 10-13 it can be seen that [HAc] 2^ [OH~] since these are 
produced in mole for mole ratio. If hydrolysis proceeds only slightly, as 
is the general rule, then [Ac~] will be about the same at equilibrium as 
it was initially and 

K H ~ [OH-] 2 /[Ac-] or [OH-] ^ 

If again c represents the molar concentration of the ion hydrolyzing 
(this time the anion), then 

[OH-] . (K u -c)l ~ [(K w lK A )e\* (10-16) 


or, solving for [H+] which is more frequently desired, 
[OH-] = 

from which by squaring both sides, rearranging, and solving for [H 4 ] one 

[H+}z*(K w K A lc)l (10-17) 

The acidity of these solutions thus varies directly with the square root of 
the ionization constant of the acid and inversely with the square root of 
the salt concentration. 

By using equation 10-17 the following tables were prepared to show 
these trends. 

TABLE 10-1. pH OF 


















of Weak Acid 


NaClO 2 


1.1 x 10~ 2 
2.0 x 10~ 4 



3.2 x 10- 8 



4.0 x 10- 10 


Why Hydrolysis Stops 

One may wonder why the hydrolysis reaction usually stops far short 
of 100% conversion. Considering cyanide ion as a typical example, the 
three equilibria concerned are, 

(a) H 2 = H+ + OH- 

(b) H 2 O + CN- = HCN + OH- 

(c) HCN = H+ + CN- 

As (b) proceeds to the right, H+ is used from H 2 O to make HCN. 
This means that (a) is displaced toward the right to furnish the H + for 
(b) and maintain the water equilibrium. Both (a) and (b) produce OH~, 
however, whose increased concentration keeps (b) from going far by the 
common ion effect. The equilibrium [H+] satisfies both K w for (a) and 
K A for (c). It is thus the build up of OH~ in this instance (and build up 
of H+ in the case of salts of weak bases and strong acids) that acts to 
limit the extent of hydrolysis. If the acid formed is exceedingly little 
dissociated, the equilibrium [H+] will be very small, so [OH~] can be 


large and hydrolysis will approach a maximum. Such is demonstrated 
in the section on sulfides. 

Hydrolysis of Salts of Monovalent Weak Acids and Weak Bases 

This is a more complex situation than those already presented, since 
both cation and anion hydrolyze. Two cases will be described. 

1. Hydrolysis if K A g* K B . If the ionization constants of the weak 
acid and weak base that are formed are about equal, derivation of a 
formula for finding [H+] is easy. Ammonium acetate is a standard 
example of this case since K A g^ K B ^ 1.8 x 1O~ 5 . 

H 2 O + NHJ- + Ac- = HAc + NH 4 OH (10-18) 

Acid 1 + base 2 = Acid 2 + base 1 
The hydrolysis constant for equation 10-18 is 

K H = [HAc][NH 4 OH]/[NH|][Ac-] (10-19) 

If the right side of equation 10-19 is multiplied by the factors [H+]/[H+] 
and [OH~]/[OH-], the terms can be separated into three products whose 
values are known: 

K n = ([HAc]/[H+][Ac-])([NH 4 OH]/[NH^][OH-]X[H+][OH-]) 

= (\IK A )(1/K B )(K W ) or 
K H = K w fK A K B ' (10-20) 

The [H+] in such solutions may be found as follows. If in equation 
10-18 we let x = [HAc] = [NH 4 OH], since the weak acid and base form 
in equimolar quantities, then if c is the original molar concentration of the 
salt, c x will be the quantity of salt ions at equilibrium: 

H 2 + NH+ + Ac- = HAc + NH 4 OH 

c-x c-x x x 


Kjf = X'XJ(c x)(c x) or 
a/(c - a) = JC /jr * (10-21) 

The H+ in the solution comes from HAc, 

^A = [H+][Ac-]/[HAc] 

and solving for [H + ] in terms of c and x 9 one gets 

[H+] = K A (x/c - x) 

This is simplified by relating the term in the parenthesis to equation 
10-21 giving 

[H+] = K A K H l (10-22) 


and relating K H to the system's individual equalibrium constants by 
equation 10-20, the final statement is 

[H+] = (K W K A IK B )* (10-23) 

Thus for salts of this type, the acidity is a direct function of the square 
root of the acid constant and an inverse function of the square root of the 
base constant. By solving for [OH~] in similar sequence one obtains 

[OH-] - K H K H * = (K W K B IK^ (10-24) 

Inspection of equation 10-23 brings forth several useful ideas: 

(a) The pR is independent of solution concentration. Term c is 

(A) If K A = K B , [H+] = K^ = [OH-] = 10- 7 M t and the solution 
^H = 7.00 

(f) KK A >K B ,pR<7 

(d) If K A < KK, pH>7 

2. Hydrolysis if K A /= K B . Equations 10-23 and 10-24 were derived 
for the case in which the two salt ions hydrolyze in equal amounts because 
KA ~ ^z?- A more general case is one in which the ion constants are not 
equal, as ammonium metaborate, NH 4 BO 2 ; K R = 1.8 x 10~ 5 , K 4 
= 6.0 x 10- 10 . Borate ion is expected to hydrolyze more than ammonium 
ion because it results in formation of a weaker electrolyte. 

H 2 O + NH+ + BOj = HBO 2 + NH 4 OH (10-25) 

This can lead to a fairly complex calculation in which three simultaneous 
equations are needed to find three unknowns, [H+], [HBO 2 ], and 
[NH 4 OH]. The problem is simplified, however, by the fact that as 
BO 2 hydrolyzes, H 2 O + BO 2 - = HBO 2 + OH", it produces hydroxyl ions 
that are used by ammonium ions, NH+ + OH~ = NH 4 OH. The latter 
reaction causes both NH+ and BO 2 to hydrolyze more than would other- 
wise be possible and the net reaction is essentially only that shown in 
equation 10 25 with [NH 4 OH] ^ [HBOJ. This effect is more significant 
than differences in ionization constants of the weak acid and base, and 
equation 10-23 can be used again. 
A typical problem follows. 

Example 10 2. HQ is a weak monoprotic organic acid that has just been 
synthesized. When NH 3 is added and the soln. coned., the pure ammonium 
salt NH 4 Q crystallizes. A 0.10 M soln. of the salt is found to have a pH of 
9.20 by ^H meter measurement. Calculate the ioniz. const. K A for HQ. 

One assumes that equation 10-23 can be used with its previously stated 


approximations. Substitution of 6.3 x 10- 10 A/ H+ (which corresponds to a 
pH of 9.2), 1 x 10~ 14 for K w and 1.8 x lO" 5 for K& one gets 

6.3 x 10- 10 = (1 x 10~ 14 A: 4 /1.8 x 10- 5 )* 

Squaring both sides and solving for K A gives 7.2 x 10~ 10 (Ans.) 
Thus HO is an acid similar to HCN in strength. 

Hydrolysis of Salts of Polyprotic Acids 

Included in this category are salts like Na 2 CO 3 , NaHCO 3 , Na 2 S, NaHS, 
Na 3 PO 4 , NaH 2 PO 4 , Na 2 HPO 4 , NaHSO 4 , etc. Problems concerning them 
can be difficult to solve since several simultaneous reactions of ionization 
and hydrolysis take place. Such calculations are considered in quantita- 
tive analysis and only a few easier examples will be illustrated. 

1. Bisulfates. The bisulfate cases are not as complicated as they 
first appear because the acid strength of HSO 4 dominates the reactions. 
For example, if one wishes to find the pH of a solution of ammonium 
bisulfate, the reactions are 

NH+ + H 2 O = NH 4 OH + H+ and 
HSO- = H+ + SOj 2 

K A for HSO- is 1.2 x 10~ 2 , which means that the dissociation of bisulfate 
is considerable. The prime effect of H+ from this source is to send the 
NHJ hydrolysis reaction to the left by the common ion principle. This 
actually occurs to such a degree that NH hydrolysis is considered 
negligible and the/>H is simply calculated as in example 7-11. 

2. Carbonates, These cases are more involved since several sets of 
equations must be carried. 

(a) Dissociations 

1. H 2 CO 3 = H+ + HCO- (10-26) 

2. HCOg- = H+ + CO" 2 (10-27) 

(b) Ionization constants 

A-I = [H+][HCO-]/[H 2 C0 3 ] = 4.16 x 10~ 7 (10-28) 

-2 = [H*][CO- 2 ]/[HC0 3 ] = 4.84 x 10' 11 (10-29) 

(c) Hydrolyses 

1. HCOg- + H 2 O = H 2 CO 3 + OH- (10-30) 

2. CO" 2 + H 2 O = HCOj + OH (10-31) 

(d) Hydrolysis constants 

a-i = [H 2 COa][OH-]/[HC03] = K W \K A . (10-32) 

-K ir lK A _* (10-33) 


Equation 10-31 is the dominant force among the reactions. This means 
that CO~ 2 is the ion which hydrolyzes more than any other and K n _ 2 will 
be larger than K R _^. Equation 10-33, therefore, is the most important 
one in the calculation of the acid or base strength of the solution. Letting 
c = molar concentration of the salt ^ [CO 3 2 ] at equilibrium, then it has 
already been shown how the following could be derived : 

[OH-] = [(K w lK A ^c}l 
[H+] = (K w K A _Jc)l (10-34) 

3. Bicarbonates. The HCO 3 ion is ampholytic and so is capable of 
giving or accepting protons : 

HCO- + OH" = CO~ 2 + H 2 O (acts as acid) (10-35) 

HCO^ + H 2 O = H 2 CO 3 + OH- (acts as base) (10-36) 

The [OH~] at equilibrium will be the difference between that produced in 
equation 10-36 and that used in equation 10-35. The quantity produced 
is equal to [H 2 CO 3 ] (since OH~ and H 2 CO 3 are formed mole for mole in 
equation 10-36 and the quantity used is equal to [CO^ 2 ] (since for every 
mole of OH~ disappearing in equation 10-35 a mole of CO~ 2 is formed), 
so one may write 

[OH~] = [H 2 C0 3 ] - [C0 3 2 ] (10-37) 

To evaluate this one refers to equations 10-28 and 10-29, which relate 
the carbonic acid and carbonate concentrations to terms that are known, 
and equation 10-37 becomes 

[OH-] = ([H+][HC0 3 ]/^) - (/WHCOjl/tH+l) (10-38) 

At equilibrium [HCO^] is about the same as it was originally when the 
bicarbonate salt solution was prepared, and this term is again called c. 
Using the identity [OH~] = tf ir /[H+] and solving equation 10-38 for the 
hydrogen ion concentration gives 

[H+] = {(K w K A ^c) + K A _,K A ^ (10-39) 

If c is not too small, K w K A ,^\c becomes insignificant and the formula for 
usual problem solving is reduced to 

[H+] = (K A _^K A _^ (10-40) 

4. Dihydrogen Orthophosphates and Monohydrogen Ortho- 
phosphates. By derivations similar to those above it can be shown that 
for solutions of salts like NaH 2 PO 4 , the hydrogen ion concentration is 
shown by equation 10-40 and for solutions of salts like Na 2 HPO 4 , the 
hydrogen ion concentration is given by 



where the /Ts are respectively the second and third ionization constants 
of orthophosphoric acid, H 3 PO 4 . It is again noteworthy that pH is 
independent of concentration and this is true for the usual range of labora- 
tory concentrations, say between 1 and 10~ 3 A/, as one can prove from 
actual measurements. 

Example 10-3. The pH of a 0.10 M soln. of Sorensen's salt, KH 2 PO 4 , is 
found to be 4.66 using the pH meter. Show that equation 10-40, which takes 
cognizance of the ampholytic character of H 2 POJ" is the correct interpretation of 
this case, whereas equation 10-17, which was derived for simple anion hydrol. 
only, does not account for the experimental data. 

Two reactions take place, hydrol. and dissoc. 

H.POr + H 2 = H 3 P0 4 + OH- 

Using equation 10-40 and the table values for the first two ioniz. constants, both 
reactions are taken into consideration : 

[H+] = [(7.5 x 10~ 3 )(6.2 x 10- 8 )]* = 2.2 x 10~ 5 M 

from which pH = 4.7, in good agreement with the measured value. (Ans.) 
Using equation 10-17 and K ^_ 1 

[H+] = [(1 x 10~ 14 )(7.5 x 10~ 3 )/0.10]* = 2.7 x 10~ 8 M 

from which /?H = 7.6, which shows that one oversimplified if he assumed that 
equation 10-17 could be used indiscriminately. (Ans.) 

5. Sulfides. These salts are important for at least two reasons: the 
first being that because K A _ l for H 2 S is so small, hydrolysis uniquely 
approaches 100%, and the second, that since sulfides are so common in 
qualitative testing it is logical to study their behavior from as many 
viewpoints as possible. 

The sulfide ion from a salt like Na 2 S will react with water to give a 
basic solution containing bisulfide ions which hydrolyze further to give 
hydrogen sulfide and more hydroxyl ions: 

1. S~ 2 + H 2 O = HS- + OH- (10-42) 

2. HS- + H 2 O = H 2 S (aq) + OH- (10-43) 
If we evaluate the hydrolysis constant for each step, 

*//-i = [HS-][OH-]/[S~ 2 ] = K W IK A ^ = 0.077 (10-44) 

K H _, = [H 2 S][OH-]/[HS-] = K w IK A .i = 1.0 X 10~ 7 (10-45) 

The magnitude of K H _^ is evidence that equation 10-42 is the important 
reaction, and since K H _ 2 is insignificant by comparison, // is permissible to 
calculate the pH of sulfide salt solutions via equation 10-44 alone. 


Example 10-4. Calculate the degree of hydrol. and pH of 0.1 M K 2 S. 

Since from the magnitude of KJJ^ we suspect extensive hydrol. it is not wise 
to use equation 10-17, which applies only when hydrol. is low. If a is the degree 
of hydrol. then the equil. concns. are 

S- 2 + H 2 ^ HS- + OH- 

(U(l-a) O.la O.la 

If a is large, (1 a) -=f=- 1, and equation 10-44 becomes, 

KJJ^ = 0.077 = (0.1 a)(0.1 a)/0.1(l - a) = 0.1 a 2 /! - a (10-46) 

Solving the quadratic gives a = 0.57. (Ans.) If one takes equation 10-43 
also into account, the hydrol. is better than 57% complete! 

From this, [OH~] ^ (0.1)(0.57) = 0.057 M, corresponding to a pH of almost 
12.8! (Ans.) 

Hydrolysis and Sulfide Solubility 

The fact that sulfide ion hydrolysis is always extensive has a pronounced 
effect upon the solubility of difficultly soluble metallic sulfides. The 
products of hydrolysis are often complex and exact problem solving is not 
attempted. For a metal sulfide, MS, the solution may contain species 
like M(H 2 0) 2 (H 2 S)+ 2 , M(H 2 O) 2 (HS) 2 , M(OH) 2 -MS, MH 2 SHS+, etc. If 
the sulfide is quite soluble, its solutions are almost as basic as NaOH 
solutions of comparable molarity, and hydroxides as well as sulfides and 
mixed compounds such as M(OH)SH have been shown to be present. 
Occasionally the hydroxide of a metal or its hydrolysis products formed in 
basic solution are less soluble than its sulfide, so they form instead of the 
sulfide when S" 2 is added. Despite the possibility of nonstoichiometric 
reactions, one can make some reasonable estimates concerning the effect 
of sulfide hydrolysis upon the solubility of sulfides. 

Example 10-5. Calculate the solub. of FeS, first from K SP alone, then by 
correcting for S~ 2 hydrol. 
The first reaction to consider is 

FeS = Fe+ 2 + S~ 2 

X X 

K SP = 4 x 10- 17 = x-x, and x = 6.3 x 10~ 9 M 

= [Fe+ 2 ] = [S- 2 ] = molar solub. of FeS (Ans.) 

However, S~ 2 is reacting with water which displaces both the following and 
the preceding reactions to the right, meaning that more FeS than calcd. goes 
into soln. : 

S- 2 + H 2 O = HS+ + OH- 

KH-I f r thk process has been shown in equation 10-44 to have a value of 0.077. 
The [S~ 2 ] from the calcn. just performed is 6.3 x 10~ 9 , and since one can assume 


it to be about 60% hydrol., the [OH~] should be 2* 3.6 x 10~ 9 A/. This 
value is lower, however, than that contributed by pure water, namely 10~ 7 A/, 
and is negligible compared with it. One may conclude that the soln. pH will 
be close to 7 and use 10~ 7 M OH~ for further calcn. : 

KH-I = 0.077 = [HS-](10- 7 )/[S- 2 ] 
[S- 2 ] = [HS-](1.3 x 10- 6 ) 

This sulfide concn. is in equil. with solid FeS and the hydrol. process as well. 
Substituting into the K SP expression [Fe +2 ] can be found : 

K 8P = 4 x 10- 17 = [Fe+ 2 ][HS-](1.3 x 10~ 6 ) 

FeS originally dissolved to give [Fe+ 2 ] = [S~ 2 ], but since most of the sulfide 
hydrol. yielding bisulfide, at equil. [Fe f2 ] r^. [HS~], which we will let be equal 

K KP = 4 x 10~ 17 = y-y(\.3 x 1Q- 6 ) 

And from this, y, which is the [Fe f2 ] and also the molar solub. of FeS, is 
5.6 x 10~ 6 M. (Ans.) 

The solubility is thus greater than originally found by a factor of over 
1000 and is actually even more than that since the further hydrolysis by 
HS~ was not considered nor was the possibility of hydrolysis of the metal 
ion, a subject treated briefly below. 

The last equation can be generalized for very slightly soluble sulfides 
of the type MS in which the first sulfide hydrolysis step is the important one : 

Molar solubility of MS = [M+ 2 ] ^ [HS~] = (K sr /L3 x 10~ 6 )* (10-47) 

Hydrolysis of Metallic Ions 

Metallic ions in solution are more or less hydrated (Table 4-1) and 
may react with the water molecules closely associated with them. This 
effect was neglected in the calculations of the preceding paragraphs. 
Again, however, probably only the first hydrolysis step is important. 
Ferrous ion may react as follows: 

Fe+ 2 + 6H 2 = Fe(H 2 0) 6 f2 = Fe(H 2 O) 5 OH+ + H+ (10-48) 

The last three substances are in equilibrium involving two equations, one 
being the water equilibrium, the other expressing the base strength of the 
pentaaquohydroxy iron(II) ion : 

H 2 = H+ + OH- 

Fe(H 2 0) 6 OH+ + H 2 O = Fe(H 2 O) 2 + OH~ (10-49) 

From K B for equation 10-49, and K w , one is able to evaluate K H for 
equation 10-48 by referring to equation 10-11: 

K H = K W \K B = [Fe(H 2 0) 5 OH+][H+]/[Fe(H 2 0)J 2 ] (10-50) 


Inspection of equations 10-48 and 10-50 shows that K n is also K A 
for Fe(H 2 O) 2 . (Why?) Generally speaking not much is done with 
calculations of this type because data for determining the extent of hydra- 
tion of ions in solution and equilibrium constants for their reactions are 
difficult to obtain. A method for approximating some values is shown in 
the next example. 

Example 10-6. A 0.1 M soln. of FeCl 2 is prepd. and the /?H, as ascertained 
with a pH meter, is 5.50. Show that K B for equation 10-49 and K J{ from 
equation 10-50 can be approximated. 

If pH = 5.50, then [H+] = 3.2 x 10~ 6 M z* [Fe(H 2 O) 5 OH+], since these 
ions are formed in equimolar quantity. If one assumes that equation 10-48 
does not proceed far (pH is not too low) then the final and initial concns. of the 
hexaaquo iron(II) ion will be about the same or, at equil.,[Fe(H 2 O) 2 ] 9^ 0.1 M. 
From equation 10-50 

K H = 1 x \0- U /K B ^ (3.2 x 10- 6 )(3.2 x 10 6 )/0.1 
from which K H g* \Q~ W and K B ^ 10~ 4 . (Ans.) 

The low magnitude of K H shows that the increase in solubility of FeS 
over that calculated from the K^* depends much more upon the hydrolysis 
of S~ 2 than the hydrolysis of Fe< 2 . 

Some Industrial Applications of Hydrolysis 

There are a number of valuable hydrolysis applications used in chemical 
industry, and chemical engineers have improved processes by applying 
equilibrium principles to their study. In the organic field these include 
the saponification of fats to make soap and glycerine, of starches to make 
sugars, of unsaturated hydrocarbons to make oxygenated products, and of 
sulfonates and halides to give hydroxylated derivatives. Inorganic 
examples include the hydrolysis of aluminum sulfate giving aluminum 
hydroxide, which clarifies water as it settles, and cheap detergent additives 
like Na 2 SiO 3 , Na 3 PO 4 , Na 2 CO 3 10H 2 O, and Na 2 B 4 O 7 - 10H 2 O, whose 
alkalinity due to hydrolysis aids in removing grease. 


1. Calculate the/>H of 0.3 M solutions of KC1, NH 4 C1, NaNO 2 , and NH 4 NO 2 . 

2. Calculate the pH of 0.2 M solutions of Na 2 S, Na 2 HPO 4 NaH 2 PO 4 , Na 2 CO 3 , 
NaHCO 3 , and NaHSO 4 . 

3. Calculate the solubility of ZnS from K SP data alone, then by also con- 
sidering sulfide hydrolysis. Explain the reason for the difference. 

4. A solution is 0.01 M in Zn+ 2 and finely powdered K 2 S is sprinkled in. 


Assuming rapid solution and mixing, does Zn(OH) 2 or ZnS precipitate first? 
See Table 17-6. 

5. Calculate the pR of 0.04 M K 2 HPO 4 by simple hydrolysis treatment, then 
by equation 10-41. Explain the reason for the difference. 

6. Explain as fully as you can this statement: "Beryllium ion is quite strongly 
hydrolyzed, hence it is impossible to prepare Be salts of weak acids, such as 
Be(CN) 2 ." 

7. A 0.1 M solution of NaOCl is tested with a few drops of thymolphthalein 
and the color produced is a very faint blue (before NaOCl bleaches the indicator). 
Show that this simple experiment allows one to approximate K A for HC1O. 

8. The /?H of MAc is 6.0. Find K B for the soluble weak base, MOH. 

9. 0.8 M NaAc is about 0.005% hydrolyzed. Use this fact to calculate the 
solution's pH. 

10. One has a weak acid that is known to be one of these: HAc, HCIO, HF, 
or HNO 2 . The pH of an 0.1 M solution of its ammonium salt is 6.3. Which 
acid is it? 

1 1 . Look over the discussion of fluoride ion in Chapter 22 and note the two 
equilibria given there. What effect will the second one have on a calculation 
concerning F~ hydrolysis? Explain. 

12. A piece of zinc is placed in a hot Na 3 PO 4 solution. It soon becomes 
covered with bubbles that are continuously being produced and expelled. When 
this gas is analyzed it is found to burn with oxygen to form only water. Explain 
with equations. 

13. (a) Explain this statement: "evidence that HX is a weak acid is the 
observation that a dilute solution of its potassium salt is basic." 

(b) It is observed that any dilution of KBr has a pH of 7.00, as has any 
dilution of NH 4 Ac. Explain each case and point out differences that exist 
between them despite the fact that both salts have the same pH. 

(c) CaCO 3 added to a solution of FeCl 3 causes Fe(OH) 3 to precipitate. 

14. One has 1 M aqueous solutions of high purity NaCN, Na 2 CO 3 , Na 2 S, 
and NaAc. Over two of the solutions very distinct odors are noted. Explain 
fully with the aid of equations and any quantitative data needed. 

15. Explain why MgS cannot be dissolved and recrystallized from water. 
How could one prepare MgS? 

16. Explain with equations how hydrolysis of both cation and anion affect 
the solubility of FePO 4 . 

17. Add two more columns to Tables 10-1 and 10-2 for % hydrolysis and 
K n . Calculate these values. How do they vary with concentration and K A ? 

18. (a) The pH of 0.2 M Zn(NO 3 ) 2 solution is found to be about 5.0. Find 
a value of KJJ for 

[Zn(H 2 0)i- 2 ] = [Zn(H 2 0) 3 OH+] + H+ 

(b) The pH of 0.1 M aluminum nitrate solution is 2.90. Calculate a value 
for the hydrolysis constant (assume only the first step is important) of 

[A1(H 2 0) 6 + 3 ] = H+ + [A1(H 2 0) 5 OH+ 2 ] (Ans. (b) 1.6 X 10~ 5 ) 


19. Give a derivation of equation 10-34 using specific ions from equations 
10-31 and 10-33. 

20. Show in a stepwise derivation that equation 10-39 is applicable to finding 
[H+] in NaH 2 PO 4 solutions. 

21. The hydrolysis constant for the first hydrolytic reaction of (Cu(H 2 O);}" 2 ] 
is about 10- 10 . Calculate the pH of 0.1 M CuBr 2 . 

22. If a = degree of hydrolysis and c = molar concentration of the salt, 
prove the following: 

(a) For NaAc, a = (K w lK 4 -c)* 

(b) For NH 4 C1, a = (K w lK B -c^ 

23. Show: (a) For NaAc, /?OH = - | log (K W 'C/K A ) 
(b) For NH 4 CI, pH = - log (K w -clK B ) 

24. Jackson Slipshod is watching a student in the quantitative analysis 
laboratory titrate a vinegar sample with standard NaOH. After thinking about 
it a few minutes J.S. announces that no accuracy is possible, since as NaAc forms 
it produces more titratable HAc by hydrolysis, therefore, one can never reach 
the end point and hence the only practical titrations are those between strong 
acids and strong bases. Has J.S. revolutionized thinking in this field? 


1. P. Van Rysselberghe and A. H. Gropp, /. Chem. Educ., 21, 96 (1944). (Sulfide 

2. K. Eiseman, /. Chem. Educ., 26, 607 (1949). (/>H and hydrolysis) 

3. L. Pokras, /. Chem. Educ., 33, 152 (1956). (Metal ion hydrolysis) 






In Chapter 1, oxidation was described as a process of electron loss and 
reduction as a process of electron gain. The oxidizing agent was defined 
as the substance gaining electrons and the reducing agent as the one 
donating electrons. In the first chemistry course one usually has an 
experiment to perform in which various metals are put in acids, and in 
solutions of each others' ions, to determine by evidence of reaction the 
relative ability of the metals to be oxidized (dissolved) or reduced (plated 
out). An "activity series" of metals is then arranged with the most active 
reductant at the top, the least active reductant at the bottom, and hydrogen 
somewhere between. In other words, a rough measure of ability to give 
and take electrons was effected. It should be possible with more subtle 
methods to make these and other redox relationships quantitative, so 
that one has data upon which to predict both the feasibility of reactions in 
unfamiliar mixtures and the extent to which the processes might go. 
These methods are known. 

Standard State, the H-Electrode and E Values 

Measurements are made on properly constituted electrochemical cells 
(Fig. 11-1) with reduction going on in one compartment and oxidation 
going on in the other. An electrode is in each compartment, and each is 
connected in an external circuit through a potentiometer or sensitive volt 
meter that gives the difference in potential (also called voltage, electro- 
motive force, or simply emf) between the two electrodes. This difference 



is a measure of the combined abilities of the reductant at one electrode 
(in one half cell) to give electrons (which travel through the external circuit) 
and the oxidant, which takes electrons at the other electrode (other half 
cell). It is necessary to compile data on such oxidations and reductions 
under definite reference conditions. These conditions of ionic concentra- 
tion, gas pressure, and temperature are called standard state, and for 
purposes here will mean ions at unit molarity, gases at 1 atmosphere 
pressure,* and a temperature of 25 C. 

A redox reaction is made up of two half reactions. Thus, with iron 
and acid 

Fe + 2H+ = Fe+ 2 + H 2 

the over-all reaction can be thought of as the sum of two half reactions. 
Iron metal is oxidized, gives electrons, and is the reductant: 

Fe = Fe+ 2 + 2e~ 

Hydrogen ions are reduced, take electrons and act as the oxidant: 

2e~ + 2H+ = // 2 

It has not been possible to accurately evaluate absolute potentials of half 
cells so they have all been referred to a standard reference electrode and 
their potentials compared to it. Since the whole cell voltage is the 
difference of the two half cell potentials, only relative potential values 
need be used in calculations anyhow. The reference electrode is the so 
called standard hydrogen electrode and it is arbitrarily assigned zero 
potential. That is, for this reaction, 

H 2 = 2H+ + 2e~ 

the standard state voltage, , is 0.00 volts. Any other half cell at standard 
state when coupled with the standard hydrogen cell will have a potential 
different from this reference cell. Since all conditions are standard and 
the H-cell contribution to the measured whole cell voltage is zero, then the 
experimentally measured voltage is E for the other half cell. In this way 
appendix Table A18 was developed. (Some values, not experimentally 
obtainable, were calculated by thermodynamical methods.) 

The H-cell consists of a solution of HC1 containing H f at unit activity^ 
into which is placed a platinum electrode coated with finely divided platinum 

* In thermodynamics the more exact measurements use unit activity of ions (molal 
concentrations corrected for interionic attractions), and \\r\\ifugacity for gases (pressures 
corrected for gas law imperfections). No attempt will be made to include these 

t This is 1.2 molar HC1 in which, due to interionic forces, H+ has an effective con- 
centration of only 1 M. Omitting the activity correction for simplicity, we will consider 
the solution to be simply 1 M in both HCl and H+. 


called platinum black. H 2 gas at 1 atm. pressure bubbles over this 
catalytically active surface. 

Some First Considerations in Setting Up Cells 

If one hooks up a hydrogen half cell with a half cell containing a copper 
electrode dipping into a 1 M cupric sulfate solution, and the two electrodes 
are connected by wires through a sensitive voltmeter, the voltmeter will 
record the whole cell potential. The laboratory set up might resemble the 
accompanying figure. 

-H solution 
x fritted, barrier 

COL electrode 
in CuSO*f solution 

FIG. 11-1. An electrochemical cell consisting of a H 2 -H + half cell and a Cu-Cu+ 2 

half cell. 

The cell compartments are separated by a fritted glass disk that permits 
only slow mixing of the solutions, although ions can travel through this 
barrier. With all concentrations at standard state, the initial voltage 
developed is 0.34 volts* and one has determined the E value for the 

* The convention for signs will be discussed in later paragraphs. 


copper half cell. If after running for a period of time, samples are with- 
drawn from each compartment and analyzed, it is found that [H + ] has 
increased and [Cu* 2 ] decreased. The half cell reactions must have been 

2<r (11-1) 


Adding these gives the over-all or whole cell reaction : 
# 2 + Cu+ 2 + 2e~ = Cu + 2H+ + 2e~ 

From reactions in equations 11-1 and 11-2 one notes that electrons are 
furnished by the hydrogen half cell, since increase in acidity also means 
production of electrons, and electrons are used by the copper half cell. 
Thus H 2 is oxidized by the oxidizing agent, Cu f2 , and Cu+ 2 is reduced by 
the reducing agent, H 2 . It will be characteristic of all cells described to 
be reversible. 

One convention adopted here is that the electric current in the external 
circuit is considered to be a stream of electrons, and since electrons are 
negative, they are attracted by the positive electrode and repelled by the 
negative electrode. Across the wire in this cell, electrons travel from the 
Pt ( ) electrode to the Cu (+) electrode. The convention to be followed 
with regard to naming the electrodes will be this : the anode is the electrode 
at which oxidation takes place (Pt, in this case) and the cathode is the 
electrode at which reduction occurs (Cu, here). In this cell, if electrons are 
leaving the hydrogen side and hydrogen ions are appearing, then negative 
ions (SO^ 2 in this example) must migrate through the permeable barrier 
from the other compartment. Electrical neutrality is simultaneously 
maintained in the copper half cell since as copper ions plate out, the 
excess of negative sulfate ions leave for the other side. 

If ions are not free to move through a barrier separating the cell com- 
partments no current will flow because of the internal resistance. This 
can be demonstrated by putting the half cells in two beakers. If a wire 
connects the electrodes, no reaction is noted. If a U-shaped tube filled 
with a concentrated solution of an inert electrolyte like KC1 is inverted 
to connect the half cells with a salt bridge that can furnish cations or 
anions on demand, current flows. If the apparatus of Fig. 1 1-1 were 
modified in this way Cl~ would go to the hydrogen side and K+ would 
go to the copper side to balance the ionic changes brought on by electron 

It is not convenient to draw pictures of cells for each discussion so 
abbreviated notation is used. The cell above may be represented as 

Pt, H+(l Af), 7/2 (1 atm)||Cu+ 2 (1 M), Cu 


The left side indicates that a platinum electrode is in contact with 1 molar 
hydrogen ions and hydrogen gas at 1 atmosphere pressure. Commas are 
used to separate different phases. The two vertical lines signify a contact 
between the two cells, as described above, which goes under the general 
name of a liquid junction. The right-hand side shows the other half cell 
Written in reverse order, cupric ions at 1 molar concentration and a 
copper metal electrode. 

Types of Half Cells 

Many different redox reactions have been studied and some elaborate 
laboratory methods used, but only four types of half cells are important. 

(1) A metallic element electrode in contact with a solution of its ions. 
This was illustrated above with Cu, Cu+ 2 . 

(2) An inert metallic element electrode in contact with a nonmetal and 
a solution of the latter's ions. This was also illustrated with Pt, J7 2 , 
and H+. 

(3) An inert metallic element electrode in contact with a solution con- 
taining both the ions of a reduced and an oxidized state of some element. 
An example of this is a platinum electrode dipping into a solution contain- 
ing both cupric and cuprous ions: Pt, Cu+ 2 , Cu+. 

(4) A metallic element electrode in contact with a mixture of a slightly 
soluble salt of that metal and a solution of common anion : Ag, AgCl, Cl~. 

One may also classify cells according >to those with liquid junctions 
(usually a salt bridge) and those without liquid junctions. The latter 
operate with a common electrolyte in both compartments. Those cells 
with liquid junctions in refined treatment involve calculations for voltage 
loss due to different ionic mobilities, resulting in a resistance at the 
junctions, but as the losses are in the order of 10~ 2 volts or less they will 
not be mentioned further. 

Oxidation Potential Conventions 

Tables of half-cell reactions and potentials are given in Appendixes 
A18 and A19. The organization and conventions of this compilation are 
listed below. 

(1) Each hak-cell equation is written with the reduced form on the 
left side and the oxidized form and electron(s) on the right. This means 
that each is written as an oxidation, hence the name "Table of Standard 
Oxidation Potentials". 

(2) Standard state for our purposes will mean ions at 1 molar concentra- 
tion, gases at 1 atmosphere pressure, and a temperature of 25 C. Voltage 
developed under these conditions is the E value. 

(3) If the reduced state in a certain half reaction in acid solution is 


better able to give electrons than is H 2 in the standard hydrogen half cell, 
then that half cell is a better reductant than H 2 and will be listed above H. 
The most powerful reducing agent (Li) is at the top of the table and poorer 
ones listed in order under it. The E values for their reactions to proceed 
as written are positive. Thus the reaction Li = Li+ + er has more 
tendency to occur than H 2 = 2H+ + 2e~. 

(4) If the oxidized state of a given half reaction is better able to take 
electrons than is H+, then it is a better oxidizing agent than H+ and will be 
listed beneath H in the table. The strongest one is at the bottom. The 
E values for their reactions to proceed as written are negative since they 
tend to go in the reverse direction. For example, from the half reaction, 
2F- = F 2 + 2e~, the large negative E value means that F 2 is a much 
better oxidant than H+, so if the fluorine and hydrogen half cells were 
coupled, the cell reaction would be H 2 + F 2 = 2H+ + 2F~. 

Conventions and Steps in emf Calculations 

Once the table of oxidation potentials is established, one has available 
a condensed collection of data from which he can predict the feasibility 
of literally thousands of redox reactions, calculate the voltage developed 
by the various combinations, determine their equilibrium constants, and 
find other values of both theoretical and practical interest. The following 
steps in solving typical problems involving the reaction equation and 
theoretical emf of the whole cell at standard state are helpful. 

(1) A whole redox reaction is made up of two half reactions. Some 
of the latter are given in Tables A18 and A 19. The two half reactions 
involved in the problem are copied from the table as they appear there. 
The form always is Reduced state = Oxidized state + electron(s), E. 

(2) It is obvious that the two half reactions comprising the whole redox 
reaction under examination cannot proceed from left to right as copied 
from the table because both are given as oxidations and one must be a 

To determine the direction in which each half reaction proceeds, so that 
their addition will give the correct overall redox reaction, one inspects 
the E values he has copied from the tables. The half cell with the larger 
, regardless of sign, will be the one that dictates the reaction direction 
since its larger indicates superior driving force for electron transfer. 
When one determines in which direction that half-cell reaction proceeds, 
he knows that the direction of the other half cell is the opposite. 

(3) We will follow the convention that a positive voltage indicates a 
spontaneous reaction, and if one is talking about a redox reaction that 
takes place, he then automatically means E for the overall reaction is 
positive. Since the overall E is found by addition of the half cell 's > 


it follows that one will always arrange for the half cell with the larger E 
to go in a direction in which the sign of E is positive. All half reactions 
and E signs are reversible.* 

(4) Consequently if the half cell with the larger value has a negative 
sign on E as copied from the table, one reverses the reaction direction and 
the sign. If on the other hand that reaction has a positive E as copied 
from the table, it proceeds spontaneously from left to right as written. 
Once this direction is determined, the other half reaction goes in the 
opposite direction, since one half reaction must produce electrons and the 
other must use electrons. Reversing any half reaction from that given in 
the tables automatically reverses the sign of E. 

Another way of thinking about this is as follows. Of the two half re- 
actions as copied from the table, the one that is more positive, or less 
negative, will be the one that is proceeding as written, since by our conven- 
tions it will be spontaneous compared to the other half cell. The latter is 
reversed together with its potential. This is equivalent to the stepwise 
reasoning given above. 

(5) If different numbers of electrons appear in the two half reactions, 
the reactions are mutually balanced by multiplying all chemical terms by 
minimum integers to give the same number of electrons in each half cell. 
This multiplication does not affect E values since they still refer to substance 
at unit activity. 

(6) After determining the electron balance, the direction of each half 
reaction, and the sign of the 's the two reactions are added to give the 
over-all cell reaction and the 's are algebraically added to give E\_ 2y 
the over-all cell voltage at standard state for half cells 1 and 2. This is the 
initial theoretical voltage (the actual may be smaller due to internal cell 
resistances) and the combined equation is the spontaneous reaction that 
takes place at that time. Since the larger E value was by these conven- 
tions made positive, J_ 2 mil always be positive and a positive emf will 
signify spontaneous reaction. One has no way of predicting, however, 
how fast this reaction will proceed, what catalyst might be needed, or how 
long it takes to attain equilibrium; but one can say that a positive E^ 2 
means that the reaction is thermodynamically feasible. Qualitatively, a 
large E_ 2 means that a reaction will probably proceed rapidly and will 
continue nearly to completion ; a small E J_ 2 signifies the opposite. 

(7) Since both of the half reactions in a given problem are written as 
oxidations, but one must be changed to a reduction to get the spontaneous 
over-all reaction (and since the change of a half reaction's direction means 
a change in sign of "), it follows that (a) if the E's have the same sign 
as copied from the table, E^_ 2 will be their difference, and (b) if the 's 

* A few listed are only theoretically reversible, but this will not affect our calculations. 


have different signs in the table, J_ 2 will be their sum. ^ 2 will be posi- 
tive in either case. 

Some Simple emf Problems 

A few examples of determining the over-all reaction equation and 
emf at standard state are given in order to illustrate the preceding general 

Example 11-1. For a cell consisting of a nickel bar in a 1 M NiSO 4 soln. 
and a zinc bar in 1 M ZnSO 4 , find the spontaneous redox reaction and the 
theoretical cell voltage at stand, state. 

From Table A18 the half reactions are: 

Zn = Zn+ 2 + 2e~ E = 0.76 

Ni = Ni+ 2 + 2e~ E = 0.25 

Both 's are positive, which means that both half reactions are spontaneous as 
written. Zinc metal, however, with its larger potential is a better red. agent 
than is nickel metal, and so electrons will leave the zinc electrode, as zinc metal 
goes into soln. and will travel to the nickel electrode where nickel ions are forced 
to accept them and nickel metal plates out. The zinc reaction dictates the 
direction of the cell reaction, and, since as written it goes left to right with 
a + , the entire nickel reaction is reversed including the E sign. Addn. 
gives the overall reaction and voltage: 

Zn ^ Zn+ 2 + 2e~ E = 0.76 

Rev. (Ni ^ Ni+ 2 + 2e~ E = 0.25) 

Zn -f Ni+ 2 + 2e~ = Zn+ 2 + Ni + 2e~ Ej_ 2 = 0.51 

This reaction proceeds if the electrodes are connected until equilibrium 
is established, and the voltage would gradually decrease to zero. One 
can deduce further that, since electrons leave zinc, that electrode must be 
negative, and, since an oxidation is taking place there, it is the anode. 
Electrons go to the positive nickel cathode where a reduction takes place. 
Since [Zn+ 2 ] is increasing in the zinc side, SO^ 2 must migrate from the 
nickel side to maintain electrical neutrality. For each SOj 2 leaving, 
however, one Ni+ 2 is being reduced to nickel metal, and so there is ionic 
balance in the nickel compartment. 

Example 11-2. One wishes to devise a test for iodide by ox. 2I~ to I 2 , since 
the appearance of the brown color of iodine (or its dark-blue reaction product if 
starch is present) would be good visual evidence of a positive test. Would 
nitric acid (in going to NO + H 2 O) be a feasible ox. agent to produce iodine 
with all substances at stand, state? 

The half reactions from appendix A18 are: 

21- = I 2 + 2e~ E = - 0.54 

NO + 2H 2 O = 4H+ + NOJT + 3e~ E = - 0.96 


Since both E's are negative it means that neither reaction tends to go as written 
but that both are spontaneous in proceeding from right to left. The stronger, 
as indicated by the larger E value, is the reaction in which nitrate reacts as an 
oxidant by using electrons, and this is the reaction that will be spontaneous. 
Because our convention is to make the E of the spontaneous reaction positive, 
the nitrate reaction will go in a reverse direction to that written in the table, 
and its E n will then be positive. The iodide reaction is forced to go as written, 
since NO^~ is a stronger ox. agent than I 2 . The least common multiple of the 
numbers of electrons in the half reactions is 6, so the iodide equation is multiplied 
by 3 and the nitrate equation by 2 to obtain electron balance. E values are 
not multiplied. Addn. then gives the over-all reaction (in which the electrons 
cancel) as well as its voltage at stand, state. Since j_ 2 is of moderate magnitude, 
and is positive for the combined spontaneous reaction as written, one may 
answer the problem's question that it is feasible at these concns. to oxidize 
iodide to iodine with nitric acid: 

61- ^ 3I 2 + 6e~ E = - 0.54 

Rev. (2NO + 4H 2 O % 8H" + 2NO^ + 6e" = Q.96) 

61- + 8H+ + 2NO 3 = 3I 2 + 2NO + 4H 2 O ?_ 2 = QA2 (Ans.) 

Example 11-3. One wishes to devise a test for formic acid, HCO 2 H, which 
is known to give bubbles of CO 2 when oxidized in acid soln. Will hydrogen 
peroxide, in being reduced to water, be strong enough to do this in I M H+ soln. 
at stand, state? What potential could theoretically be developed? 

The half reactions from Table A18 are 

HCO 2 H (aq) * = C0 2 + 2H^ + 2?- E = 0.20 

2H 2 = H 2 O 2 + 2H+ + 2e~ E = - 1.77 

The larger E value is from the hydrogen peroxide half cell so it will be 
spontaneous in the direction in which E is positive, namely the reverse of that 
given in the table and copied above. This means that with H 2 O 2 taking electrons 
in its role as oxidizing agent, HCO 2 H must give electrons as a reducing agent 
and its reaction goes as written : 

HCO 2 H ( a q) r^ CO 2 + 2H+ + 2e~ E = 0.20 

_ Rev. (2H 2 O % H 2 O 2 + 2H+ + 2e~ E = 1.77) 

HCO 2 H (a q) + H 8 O 2 = 2H 2 O + CO 2 ^ 2 = 1.97 (Ans.) 

The large Ej_ 2 shows that the reaction oxidizing the acid would be quantitative. 

The Calomel Electrode 

While the hydrogen electrode is the zero potential standard it is an 
experimentally aggravating device to use. The platinum black surface 
must be uniform, clean, and continually resaturated with H 2 . Its use is 

* (aq) means an aqueous solution of the substance, to differentiate it from its an- 
hydrous form. 


limited to solutions that do not contain substances that are reduced by 
H 2 or oxidized by H+. Volatile substances like CO 2 must also not be a 
part of the reaction, since they are removed with the H 2 excess. Even at 
best, the H-electrode may be erratic and slow in its response. 

For these reasons, other standard reference electrodes have been 
developed. One of the best is the normal calomel electrode which gets its 
name from the common name for Hg 2 Cl 2 , calomel, and also contains 
1 M KC1 and mercury metal. These electrodes are easy to prepare and 
give repetitive results. The reaction is, 

2Hg + 2C1- = Hg 2 CI 2 + 2er E = - 0.28 

A typical design is shown in Fig. 1 1-2. 

The Nernst Equation and Equilibrium Constants 

Reactions and cell emf's so far considered have been for processes at 
standard state. The question naturally arises as to what happens at other 
concentrations and temperatures? A formula developed in 1889 by 
W. Nernst supplies the relationships: 

ceii = ?- 2 - (1-98 x 10- 4 7 log Q (11-3) 

= the emf of the whole cell at conditions other than standard, 
_ 2 = emf of the whole cell at standard state as already illustrated, 
T = the absolute temperature at which the cell operates, n = the number 
of electrons involved in either the oxidation or reduction when the half 
cells are mutually balanced, and Q = the product of molar concentrations 
(more accurately, activities) of reaction products divided by the product 
of molar concentrations of reactants, each term raised to a power corres- 
ponding to its coefficient in the balanced over-all equation. Thus Q has 
the form of the equilibrium constant of the reaction, but the terms are 
usually for experimental conditions, not equilibrium conditions. If the 
temperature is 25 C, 

= E_ 2 - (0.059/n) log Q (11-4) 

All calculations henceforth will be made for 25 unless otherwise specified. 
If the terms comprising Q are equilibrium concentrations, then Q = K, 
the equilibrium constant. At equilibrium, however, there is no net electron 
or ion transfer, so the cell voltage is E cc ii = and the Nernst equation 
can be solved for the equilibrium constant: 

= _ 2 - (0.059/rt) log K 

_ 2 = (0.059/Ai) log K (1 1-5) 



log K = E_ 2 /0.059 (11-6) 


K = lO 17 "^ 1 - 2 (11-7) 

By proper combination of half cells and the measurement of concentrations 
and whole cell voltage, many equilibrium constants of interest like K S1 >, 
K H , K A , K B , and K im have been determined. Knowing these constants, 
the chemist can use equation 11-6 in reverse fashion to determine equili- 
brium concentrations, such as [H+] (therefore the pH of an unknown 
solution) or the quantities of other ions left after reactions, and also to 
see whether or not redox reactions are at least theoretically quantitative. 

Example 11-4. In the Reinsch test described in Chapter 16, one displaces 
"heavy metals" from soln. by using a superior reducing agent, copper metal. 
A copper wire is placed in a small amount of 10 3 M Hg+ 2 soln. Calculate K 
for the reaction and the equil. [Hg+ 2 ]. From this information comment on 
the test sensitivity. 
The half reactions to look up from Table A18 will be those concerning 

Cu Cu +2 and Hg Hg f2 , and when written and the mercury reaction and its 

E sign reversed, we have: 

Cu ^ Cu+ 2 + 2e~ E = - 0.34 

Rev. (Hg 7 Hg+ 2 + 2e~ E = 0.85) 

Cu + Hg+ 2 = Cu+ 2 + Hg % j_ a = 0.51 

The desired reaction, that is, displacement of the heavy metal by copper metal, 
is the spontaneous one, proceeding with a respectable voltage. 
K is calculated using equation 11-7: 

K = [Cu+ 2 ]/[Hg+ 2 ] = 10 (17)(2)(0 - 51) = 2.0 x 10 17 (Ans.) 

As explained in the discussion on K Ht > in Chapter 8, the concns. of solids are 
regarded as constants and do not appear. Since K is large, we may assume the 
reaction is quant, and that almost all the Hg has been reduced, making the 
equil. [Cu+ 2 ]^ 10~ 3 Af (since 1 Cu + 2 if formed for every Hg^ 2 disappearing 
and 10~ 3 M was the orig. [Hg+ 2 ]). Therefore, 

K = [Cu+ 2 ]/[Hg+ 2 ] = 2.0 x 10 17 = 10' 3 /[Hg+ 2 ] 

from which [Hg+ 2 ] at equil. = 5.0 x 10~ 21 M (Ans.) 

The reduction of Hg+ 2 is essentially complete, and the test is a sensitive one, 

capable of detecting as little mercury for which one can visibly find displacement 


Deductions from the Magnitude of 1 _ 2 

If E\ . 2 of a particular cell is near zero volts by virtue of selection of half 
cells whose spontaneous reaction potentials are about alike, the standard 



state concentrations are close to those of equilibrium. Relatively minor 
concentration changes can, by the mass action effect, cause the spontaneous 
reaction to be the reverse of the one originally found. This is the case 
when Eoeii comes out to be negative. On the other hand, if _ 2 i l ar g e 
then K is large and there is no likelihood of equilibrium conditions being 
reached until some concentrations become vanishingly small. 

Electrochemical Determination of pH 

One way of finding the pH of a solution is to couple a calomel and a 
hydrogen cell as shown in Fig. 1 1-2. 








FIG. 11-2. A calomel-hydrogen electrode system for finding /?H. 

The cell data is 
Rev. (2Hg + 2C1 

- 2H+ 

Hg 2 Cl 2 = 2Hg + 2C1 

E = 0.28) 

= 0.00 
J_ 2 = 0.28 


If everything but [H f ] (the solution's unknown acid strength) is at standard 
state, then Q simply is [H f ] 2 . The Nernst equation then is 

0011 - 0.28 - (0.059/2) log [H+] 2 
which reduces to 

E wn = 0.28 + 0.059 pH (1 1-8) 


yH = 17(*ceii-0.28) (11-9) 

This means that if one measures the experimental cell voltage, the solution's 
pH is found in one simple calculation. If the solution contains a weak 
acid such as HY at known concentration r, the pH can be converted to 
[H { ], which will be equal to [Y~] and the ionization constant follows since 
all other terms are known in K A = [H+][Y~]/c. See example 7-5. 

Electrochemical Determination of K Nl , and K tn8 

Correct construction can result in a cell for which the observed cell 
voltage is a measure of the solubility of a slightly soluble salt or of the 
dissociation of a complex ion. This is done by arranging the other cell 
components to be solids or ions at standard state, so they conveniently 
cancel from the calculations. 

One can also calculate equilibrium constants from values given in 
Table A 18 without making any lab measurements. This is illustrated in 
example 1 1-8. See also problems 7 and 8 on p. 186. 

Example 11-5. Calculate the instability constant of Ag(S 2 O 3 )^ 3 from 
data given in Table A 18. 

3 ] 

the cell data selected must include these three ions. The two half reactions 
which do this are 

Rev. (Ag *=? Ag+ + e~ E J = + 0.80) 

Ag + 2S 2 (V e-> Ag(S 2 3 ) 2 3 + <r E J = - 0.01 

Ag + 2S 2 3 2 = Ag + Ag(S 2 3 ) 2 3 E?_ 2 = 0.79 

The solid Ag terms are constant, so 
Q = [Ag^Oa 

At equil. " C eii is zero so the Nernst equation is 

= 0.79 - (0.059/1) log (l/jr ijw ) = 0.79 + 0.059 log K ins 
from which we get an answer checking with the value from Table 1 5-1 : 

K in8 = 10- 13 - 4 = 4 x 10~ 14 (Ans.) 


Equilibrium Constants for Redox Half Reactions 

By methods already given, one can calculate the equilibrium constant 
associated with each of the half reactions given in the tables. It is but a 
simple step further to combine any two of them to get K fq for a whole 

Example 11-6. One has the cell Pb, Pb+ 2 ||Cu +2 , Cu. Find K for each side, 
and Kfq for the spontaneous over-all reaction. 
From Table A18 we copy 

Pb = Pb+ 2 + 2e ~ E = 0.126 

Cu = Cu< 2 + 2e~ E = - 0.337 

By equation 11-7, the equil. const, for the Pb half reaction is 

/r I>b = [Pb+ 2 ]/[Pb] = [Pb+ 2 ] = 10< 2 >< 17 >< - 12 > = 1.95 x 10 4 

and for the Cu reaction is K Cu = [Cu+ 2 ]/[Cu] = [Cu+ 2 ] = io< 2) < 17) <-- 337) 

= 3.55 X 10- 12 (Ans.) 

If one now combines the two half reactions from Table A 18 as shown in 
previous problems, he finds the spontaneous reaction and over-all voltage is 

Cu+ 2 + Pb = Pb+ 2 + Cu E- 2 = 0.463 

and the equil. const, is [Pb +2 ]/[Cu 42 ]. But these values were just calcd., so 
K eq = 1.95 x 10 4 /3.55 x 10' 12 = 5.50 X 10 15 (Ans.) 
The result can be checked by the method used to find K eq in previous problems. 

Potential Diagrams 

Speculation on reaction feasibilities can be facilitated by what are known 
as potential diagrams. These are condensed versions of half-cell data that 
summarize potentials and valence states of a given element. By means of 
the diagrams one can quickly see how interconversions among the valence 
states can be effected. In addition, if one knows something of the chem- 
istry of these states, he has an excellent summary of the element's chemistry. 

Suppose we are interested in devising a separation between uranium 
and plutonium, and cell study in acid solution has given us data to con- 
struct the following diagrams. (Reading right to left, the E values are 

0.61 TT a A -0.62 + 0.05 4 .3 


n 2.03 ^ , 

Pu Pu + 


Two separation paths seem worth investigating. Given a solution of 
U+ 3 and Pu 43 , a very mild oxidizing agent can oxidize U 43 -> U 44 , a spon- 
taneous reaction anyhow, whereas Pu 43 does not go to Pu 44 because its 
large negative potential means it is difficult to oxidize to the + 4 state. 
If an anion, X~, is then added, such that UX 4 is say insoluble, volatile, or 
perhaps extractable in some immiscible solvent, while PuX 3 is not, separa- 
tion can be made. 

Another route is from a solution containing U 44 and Pu 44 . The 
uranium is not difficult to oxidize to UO 4 , 2 , whereas the large -1.04 value 
for the conversion Pu 44 -> PuO 4 2 , precludes the latter happening simul- 
taneously with a properly selected oxidant. The resulting solution con- 
tains only UO 42 and Pu+ 4 now, and as above, a precipitant, etc. is then 

Some Practical Applications 

The discussion thus far has concerned theoretical electrochemistry as 
used by chemists for a quantitative approach to redox reactions. The 
same principles apply wherever electrochemical phenomena are found. 

1. Electroly ,/*\ In a cell in which electrolysis is taking place, electrode 
polarity is dictatH by an external source of potential, such as a battery of 
the type already described. The battery can be thought of as an electron 
"pump," causing an electron deficiency at the electrolysis cell's anode 
and an electron excess at the cathode. In the electrolysis cell, positive 
ions, therefore, move toward the negative cathode and may gain electrons 
and be neutralized, as Cu+ 2 -f 2e~ = Cu. Negative ions move toward 
the positive anode and may give up electrons and be neutralized, as 
2Br~ = Br 2 + 2e~. Thus, electrons do not move through the solution 
but electrical conduction occurs by ionic migration. 

Two principles governing electrodepositions were discovered by 
M. Faraday in 1834, and are known as Faraday's Laws: 

(a) The weight of a substance produced at an electrode is directly 
proportional to the quantity of electricity passing through the cell. 

(b) Identical numbers of equivalents of different substances can be 
liberated at electrodes by the passage of a given quantity of electricity. 

The quantity of electricity liberating a gram equivalent weight is called 
afaraday. It is 6.02 X 10 23 electrons (Avogadro's number) and equals 
96,500 coulombs. A coulomb is that quantity of electricity which passes a 
given point in a conductor in one second when the current is one ampere. 
Thus C = It, where C is coulombs, t is time in seconds, and / is current in 
amperes. The equivalent weight of an ionic substance undergoing 
reaction at an electrode is the ionic weight divided by the number of 
electrons per ion needed for its oxidation or reduction. A faraday could 


therefore plate out 107.88/1 grams of silver, 63.54/2 grams of copper, and 
26.98/3 grams of aluminum. 


Example 11-7. Find the Ibs of Mg and the s.c. ft 3 of C1 2 produced by 
electrolysis of a molten MgCl 2 bath using 100 amps for 1 hr. (see Mg, Chapter 19). 
The electrode reactions are : 

Mg+ 2 -f 2e- = Mg (cathode) 

2C1- = C/ 2 + 2e- (anode) 

Thus the equiv. wts are Mg/2 = 12.2 and CI 2 /2 = 35.5. The number of 
coulombs is (100 amp)(3600 sec) = 3.6 x 10 5 , and the number of faradays is 
3.6 x 10 5 /9.65 x 10 4 = 3.74. Since one faraday deposits a g equiv. wt and 
since 454 g = 1 Ib, the Ibs produced will be (3.74)(12.2)/454 = 0.100 Ib of Mg 
(Ans.) and (3.74)(35.5)/454 = 0.293 Ib of C1 2 . A Ib mol. wt of any gas at s.c. 
occupies about 359 ft 3 , so the vol. produced is (0.293)(359)/71.0 = 1.48 ft 3 of 
C1 2 . (Ans.) 

Industrial electrochemistry includes preparation and/or purification of 
metals by reduction from aqueous solution (Cu from CuSO 4 ) and from 
molten salt baths (Al from A1 2 O 3 + Na 3 AlF 6 ), as well as preparation of 
chemicals by oxidation (H 2 S 2 6 8 from H 2 SO 4 + H 2 O, KMnO 4 from 
K 2 MnO 4 , etc.). 

2. Corrosion via electrochemical mechanisms. The analyst is 
sometimes asked to analyze corrosion products as a means of determining 
how to protect metals. Diagnosis is complicated by the possible presence 
of several paths by which electrochemical deterioration can occur. 

(a) Galvanic cells. Two dissimilar metals in contact with each other 
and a solution constitute a short-circuited cell of the type given in the 
first part of this chapter. As one example, a brass fitting on an iron water 
pipe sets up the reaction Fe + Cu+ 2 = Fe+ 2 + Cu, as determined by the 
oxidation potentials. Air oxidation and hydrolysis of Fe+ 3 gives a precipi- 
tate of Fe(OH) 3 , further displacing the reaction to the right. 

Galvanizing and cadmium plating and the use of attached anodes of 
Mg are ways in which the same principle is used to protect iron at the 
expense of sacrificial metals. Under nonlaboratory conditions, such as a 
tank buried in moist soil, a ship in salt water, or new alloys in use, the 
usual metal activity series and table values of oxidation potentials are 
often of little value, and the experimenter must determine the characteristics 
of each new situation. 

(b) Concentration cells. These are usually systems in which the same 
type of electrodes are in contact with a solution of varying composition; 
for example, an iron well casing penetrating two salt water pools of 
different concentration. 

Metal ions go into solution with the aid of a force called solution 


pressure and are opposed from doing so by the force of osmotic pressure 
manifested by ions already in solution. Although these pressures are 
difficult to measure, one may use their concept to explain that in the more 
concentrated solution, the high osmotic pressure hinders further ionic 
production, hence the metal corrodes more in the more dilute solution. 

(c) Differential temperature and differential strain cells. A high 
temperature accelerates ionic movement and gas release and speeds 
corrosion. Corrosion is also favored in areas in which metal is strained, as 
nonannealed areas and where stresses develop in use. 

(d) Differential aeration cells. These cells often consist of a metal in 
contact with two solutions differing only in quantity of dissolved air, and 
it may be surprising to learn that corrosion primarily takes place in the 
solution containing the lesser oxygen concentration. Consider trying to 
dissolve copper in hydrochloric acid: 

Rev. (Cu ^ Cu f2 + 2e~ = 0.34) 

//a -' 2H * + 2<r /T =0.00 

// 2 = 2H ' + Cu __ 2 ^O34 

One calculates the equilibrium constant to be 

K= [H'] a /[Cu ]/>, = 3.6 X 10 11 

The tendency for copper metal to reduce protons and form hydrogen gas 
is therefore slight, but since this is an equilibrium process one may 
assume that a thin film of H 2 forms on the Cu to establish early equilibrium. 
The copper is said to be insulated by the H 2 or is polarized. If oxygen 
is readily available to depolarize the electrode by reaction with the 
hydrogen, an electron deficiency is imminent on metal where the oxygen 
concentration is high. This exerts an electron demand on areas which are 
not depolarized so the latter give up electrons, become anodic, and metal 
dissolves there. By this mechanism, corrosion is prevalent in that part 
of the system containing comparatively little oxygen. Water systems are 
mechanically de-oxygenated by vacuum, or chemically, using Na 2 SO ;j , 
NaNO 2 , or hydrazine hydrate, (NH 2 ) 2 H 2 O. 

(e) Impressed current cells. An outside source of current such as a leak 
from dc machinery may force electrons to flow away from a metal surface 
such as a pipe, causing metal to deteriorate at that point. Water or water 
plus dissolved oxygen is the recipient of electrons, the former giving 
OH- + H 2 as products, the latter giving only OH . Ferric hydroxide is 
the final product and its deposition may or may not yield a protective 
coating to the metal surface. Repairing or shielding machinery to 
minimize current leaks stops this corrosion in its first step. 



1. For each reaction calculate (a) the spontaneous standard state reaction 
equation (b) "i_ 2 (c) the direction of electron flow in the external circuit (d) the 
polarity of the electrodes (e) K: 

(1) Cd + 2H+ = // 2 + Cd+ 2 

(2) Sn + 2H+ = 7/2 + Sn+ 2 

(3) Cd+ 2 + Sn = Sn'- 2 + Cd 

(4) C1 2 4- 21- = T 2 -f 20- 

(5) 0> + 4H+ + 4Cr+ 2 = 4Cr+ 3 + 2H 2 O 

2. Calculate the items given in problem 1 for each of the indicated cells at 
standard state : 

(1) Pt, Fe+ 2 , Fe+ 3 1| // 2 , H+, Pt 

(2) Fe, Fe+ 2 1| Cd+ 2 , Cd 

(3) Mg, Mg+ 2 || Br-, Br 2(aq) , Pt 

(4) Ag, Ag+ || Pb^ 2 , Pb 

(5) Pt, Sn+ 2 , Sn+ 4 1 1 Cu+ 2 , Cu+, Pt 

3. The Daniell cell consists of a copper bar in a solution of cupric sulfate and 
a zinc bar in a solution of zinc sulfate. (a) Draw a picture of the cell. Calculate : 
(b) El- 2 (c) K(d)E c ^ } if [Cu+ 2 ] = 10~ 4 Mand [Zn< 2 ] = 2 M (e) E ce ii if [Cu f2 ] 
= 3 M and [Zn+ 2 ] = 10- M (f) C cii if [Cu+ 2 ] = 4 M and [Zn+ 2 ] = 4 M. 
What will be the effect if to the cell at standard state one adds: (g) H 2 S to the 
copper side? (//) NH 4 OH to the zinc side? (/) heat to raise the temperature from 
25 to 50 C? (/ ) an equal excess of NH 4 OH to both sides? 

4. (a) Why does E C eii = at equilibrium? 

(b) Why does E C eii change with concentration changes ? 

(c) What is meant by standard state? 

(d) Why does Q = K at equilibrium? 

(e) Which cell would theoretically give the greatest emf at standard state? 
Which the least? 

5. One has the cell Pt, H 2 (latm.), H^(M)||H+ (?), H 2 (latm.), Pt and 
experimentally finds " ce ii = 0.21. Find [H+ ?] and pH. 

6. A calomel cell and a hydrogen cell are coupled. All substances are at 
standard state except the acid solution, which consists of 0.1 M HZ, a weak 
acid. The measured cell voltage is 0.575. Calculate (a) pU (b) K A for HZ. 
(Am. (a) 5.0 (b) 10~ 9 ). 

7. Find the K SP of AgBr from 

Ag + Br- = AgBr + e~ E = - 0.06 

Ag = Ag+ + e~ E = - 0.80 

8. Find the K 8P of PbCl 2 from the proper E values. 


9. Find in Table A18 one example of each of the four types of half cells men- 
tioned early in this chapter. Choose examples not previously discussed. 

10. Find the spontaneous reactions and K's to support your answers for the 

(a) It is stated in the group 3 cation discussion that some FeS may form 
when H 2 S is added to acidified Fe+ 3 solution due to the reducing effect of H 2 S 
on ferric iron. Does this reaction appear to be quantitative? 

(b) Will a spiral of silver wire wrapped around an iron pipe in moist soil keep 
the iron from corroding? 

(c) Will Fe+ 3 be reduced to Fe+ 2 by iron metal? 

(d) It is reported that one can test for ozone by bubbling it through an acidic 
Mn+ 2 solution, whereupon purple MnO^ is formed. Is this feasible? 

11. One has the two cells shown in Fig. 11-3. In each set up, tell which 
end of the iron horseshoe corrodes more rapidly and why. 

o i /y 


FIG. 1 1-3. Two schematic experimental corrosion cells. 

12. (a) On the basis of hydrogen depolarization, explain why Cu dissolves in 
HNO 3 yet H 2 is not given off. Why does Cu dissolve in HC1 4- H 2 O 2 but not 
in HC1 or H 2 O 2 alone? 

(b) It is observed that a ship rusts badly at several spots well below the 
water line, where air is not readily available. Explain. 

13. Balance the following half reactions and obtain the over-all reaction. 
Will this reaction proceed furthest at low or high pH (neglecting possible side 
reactions) ? 

Mn+ 2 + H 2 = MnO<f + H*- + e~ 

H 2 2 = H+ + 2 + <r 

14. Draw a simple diagram to illustrate one practical example of each of the 
five types of cells mentioned in the section on corrosion. Indicate where 
corrosion takes place and the electron flow in each. 

15. Suppose you are working on an atomic energy program and are given 
some strips of neptunium for the determination of E for 

Np = Np+ 2 + 2e~ 

Discounting handling problems due to its radioactivity, draw a set up you might 
use and explain. 


1 6. A man has a large gold dental bridge in his mouth. He touches it with an 
aluminum spoon and experiences what he describes as an electric shock. 

17. From the reactions 

Ag + 2NH 3 = Ag(NH 3 )+ + e~ E" = - 0.373 

Ag = Ag+ + e- = - 0.799, 

show that K IH i for diammine silver(l) ion is about 6 x 10~ 8 . 

18. Show by calculation of the spontaneous reaction and K that magnesium 
will act as an anode and protect lead from corrosion. If these metals were in a 
laboratory cell with [Pb+ 2 ] and [Mg+ 2 ] initially 0.1 A/, calculate the equilibrium 
lead ion concentration. 

19. In organic qualitative analysis, one matches an unknown with a known 
sample as proof of their similarity by means of a mixed melting point. This 
entails intimately mixing the samples and slowly heating the mixture. A pure 
compound melts over a short range (1 C), whereas a mixture (except eutectics) 
melts over a wide range. In somewhat analogous fashion the following is 
suggested as a qualitative testing method for metallic samples: "Connect the 
known metal and unknown metal to the terminals of a milliameter and immerse 
the metals in dilute sulfuric acid solution." Explain what one would observe 
and why. 

20. J. P. Slipshod describes the electrolysis of aqueous sodium chloride using 
Pt electrodes this way: "Cl~ goes to the anode, takes electrons, and becomes 
C1 2 gas which reacts with H 2 O forming HC1O. Na+ goes to the cathode and 
changes to Na atoms which immediately react with H 2 O giving NaOH + H 2 ." 
What is your version of this process? 

21 . A large excess of neutral CuSO 4 solution is electrolyzed at Pt electrodes for 
15 minutes, using 3 amperes. The anode reaction is 2H 2 O = 4H+ -f (\ + 4c . 
Find (a) the number of faradays and coulombs used (h) g of Cu and liters of O 2 
produced, (r) g equivalents of W produced (d ) final solution /;H. 

22. A silver coulomcter contains Ag electrodes in an AgNO :1 solution. 
Electricity is passed through the cell and the weight of Ag gained by the cathode 
or lost by the anode is determined and used to calculate the quantity of electricity 
involved. If after 10.0 minutes, 1 .341 6 g of Ag are lost by the anode, what was 
the current strength in amperes? (Ans. 2.0 amp) 

23. You are asked to go to the laboratory and, using simple equipment, to 
determine the number of coulombs in a faraday. Explain your method and 
include a labelled drawing of the experimental setup. 

24. Recalculate K for each of the reactions in problem 1 using the method of 
Example 1 1-6. 

25. Which will fail sooner in use and why: copper plates riveted together 
with iron rivets or iron plates riveted together with copper rivets? 

26. Given the following hypothetical potential diagram for metal M, in acid 

MO ^-^- 


Explain how one can deduce these conclusions: (a) M dissolves in dilute acids, 
releasing H 2 . (/>) M+ 2 tends to disproportionate, (c) MO 4 2 is an oxidant 
similar to permanganate in strength, (d) Br 2 is powerful enough to oxidize M+ 4 
to MO+ 3 . (e) Addition of metallic M to a Cu +2 solution yields a mixture of 
Cu and M -*. 


1. W. T. Hall, J. Chem. Educ., 21, 403 (1944). (emf series) 

2. W. F. Luder and A. A. Vernon, J. Chem. Educ., 22, 6? (1945). (emf conventions) 

3. R. K. McAlpine, /. Chem. Educ., 23, 301 (1946). (Some oxidmng agents) 

4. R. B. Spacht, J. Chem. Educ., 23, 195 and 253 (1946). (Corrosion) 

5. J. A. Timm, J. Chem. Educ., 24, 160 (1947). (Electrolysis) 

6. A. W. Davidson, /. Chem. Educ., 25, 533 (1948). (Electrochemistry) (See also 
the papers following in that issue) 

7. W. F. Kieffer, /. Chem. Educ., 27, 659 (1950). (Activity series) 

8. A. E. Harvey and D. L. Manning, J. Chem. Eiluc., 28, 527 (1951 ). (Experimental 
oxidation potentials) 

9. R. J. Gettens, ./. Chem. Educ., 28, 67 (1951). (Corrosion products) 

10. L. G. Sillen, J. Chem. Eiluc., 29, 600 (1952). (Redox diagrams) 

11. S. 1. Miller, J. Chem. Educ., 29, 140 (1952). (Combining half reactions; 

12. R. M. Burns, /. Chem. Educ., 30, 318 (1953). (Corrosion reactions) 





Fifty years ago, the chemical analyst had only a modest number of 
tools at his disposal. A man with some knowledge of chemical reactions, 
a small supply of chemicals and glassware, a few books, and a chemical 
balance was equipped about as well as the next chemist for the performance 
of chemical analyses. This is not true today. Although much of the 
fundamental theory of analytical separations is the same, the number of 
methods and their sensitivity has increased greatly. Two important fields 
where advancement has been most fruitful are instrumentation and the 
use of organic reagents. The analyst should know something of these 
developments, their application, and potentiality. Scarcely an issue of an 
analytical journal appears now without a majority of articles on these two 
topics. While the manipulation of instruments is reserved for more 
advanced study, a little familiarity with organic analytical reagents is de- 
sirable at this time, since the use of several of them will be described in the 
laboratory directions. 

Advantages of Organic Reagents 

1 . They can be more specific than convention il reagents. Because of 
this property they can frequently be used to detect a single type of ion in a 
complex mixture. The time may soon be at han ' when enough is known 
about organic reagents and enough of them are a ailable so that a relative 
novice may run qualitative analyses with their v e alone and circumvent 
more laborious inorganic separations and tests. his prospect particularly 
appeals to students. 



2. They can be more sensitive. Classical semimicro-methods of 
qualitative analysis will usually not detect much less than 100 gammas 
(lOOOy = 1 mg) of a substance under specific test conditions, although a 
few are better than that. Organic reagents, on the other hand, are capable 
of detecting absolute amounts of desired substances below lOy and in 
certain cases, as little as 10 3 y. This rivals the spectrograph as a means 
of detecting traces of metals. Concentrationwise, one expresses the quan- 
tities usually as y/ml =.- mg/liter = parts per million (ppm), and concen- 
tration limits of many tests using organic compounds are well below 
1 ppm. The sensitivity of these methods is due to the formation of a 
highly colored material which is discernible at low concentration, or to 
the production of a precipitate which, due to the size of the organic 
molecules involved, has a molecular weight several times that of the 
substance whose presence it indicates. If large samples are available for 
concentration, then almost no lower limit may be put on the percentage 
of a constituent that can be found. 

3. They can be adapted to quantitative analysis. The organic precipi- 
tates are generally insoluble in water at a given /?H and may be separated, 
dried, and weighed. Those reactions giving colored, soluble products 
frequently show a regular change in color intensity with concentration 
of the substance under test and so are adaptable to accurate comparison 
with colors representing known concentrations. This is the basic idea in 
that branch of the science called colorimetty. The color matching may be 
done visually or by use of instruments like the colorimeter or spectro- 
photometer designed for the purpose. Many thousands of methods in 
colorimetry have been reported and are used daily in clinical laboratories, 
hospitals, etc., where production with accuracy is stressed. Some examples 
are given of this in the preliminary laboratory exercises and in the special 
experiments. (See special experiment 3.) 

Disadvantages of Organic Reagents 

1. The application of organic compounds is still somewhat new and 
many structures of products and mechanisms of reactions are not known. 
This makes discussion of tests limited and always gives an uncertain 
feeling, since one does not know what factors must be regulated. As with 
any new reactions, interferences are not always known and may not have 
been studied or reported. And from the teacher-student viewpoint, little 
chemistry is to be taught unless fairly definite concepts are dealt with and 
these are lacking in some empirical tests. 

2. Organic reagents are many times not too stable toward certain 
solvents, /?H, oxidi/.mg and reducing agents, and temperature changes. 
They may, under these influences, be destroyed or render false colors 


and/or precipitates. For this reason, along with the unknown sample, 
one should always run a known and a blank sample to check the reagent 
itself. Some reagents must be made just prior to use because of their 
sensitivity, particularly toward air oxidation. 

3. The sensitivity of the reagents may be so great that extraordinary 
precautions are needed to insure their proper functioning and because of 
the time involved, other tests may be preferable. Sometimes the useful 
color range is in the order 0-2 ppm, and for quantity estimation great 
sample dilution is necessary. There are reagents, for instance, that readily 
detect copper in water condensed in copper tubing, give vivid tests for 
zinc from galvanized pipes, and show metallic impurities in filter paper. 
They are too sensitive for student purposes. 

Kinds of Reagents 

A number of classifications of organic analytical reagents are possible. 
The four given below are based upon mode of reaction with the test ions. 
Laboratory exercises are later described using each reagent. 

The organic chemist needs to write structural rather than molecular 
formulas for the compounds he describes because the latter are not 
adequate. This simply means that dots arc used to show unshared 
electrons, lines indicate covalent bonds, and the important elements in 

organic chemistry are then given as C , H , : O , N , and : S . 

About 800,000 organic compounds are known. This means there are 
approximately three times as many carbon compounds known as com- 
pounds of the other 100 elements. The diversity is due to the ability of 
carbon atoms to form strong covalent bonds not only with other kinds 
of atoms, but also with themselves, thus building up molecules con- 
taining chains and rings of atoms in great variety. In addition, carbon- 
carbon covalent bonds may be single, double, and triple as related in 
Chapter 3. 

Type A. The reagent forms a chelate, or inner complex, with the sub- 
stance being tested due to the presence and proximity of groups capable of 
coordination. The resulting substance is either colored or insoluble or 

Chelates have been described in Chapter 4. When the electron donating 
atoms O, N, and/or S are part of the organic compound and 5- and 6- 
membcred rings can form by incorporating the inorganic ion as part of an 
expanded structure, chelation may occur. The increase in molecular 
weight usually means a decrease in solubility. 


L Dimethylglyoxime for Ni+ 2 . 

H 3 C\ |-' : 

H 3 C C=N 6 H C ^x, / C ?.~~ N 

2 | " " -fNi^ = H 3 C-CV Ni ,C CH 3 + 2H' 

H,C C=N a- H % N-Q *N --.c:' 

F .1 CH 3 

H - O B . 

The chclate is not appreciably water-soluble and is red colored. It has 
the possibility of closing four rings by application of three bond types: 
covalent (lines), coordinate (arrows), and hydrogen (dots). The hydrogen 
bonds are weak, but the four links to Ni arc strong, and the chelate is 
stable. The test will detect 0.3 ppm Ni+ 2 . Sec cation test, 17-11. 

2. Dithizone (diphenylthiocarbazone) for Heavy Metals. This re- 
agent can be used as a screening agent for solutions suspected of containing 
any of these heavy metal ions: Bi^ 3 , Cd+ 2 , Co' 2 , Cu^, Cu+ 2 , Au 1 , Iiv* 3 , 
Fe^ 2 , Pb* 2 , Mn+ 2 , Hg^ 2 , Hg+ 2 , Ni+ 2 , Pd 12 , Pt+ 2 , Ag ( , TP, T1+ 3 , Sn f2 , 
and Zn+ 2 . If under specified experimental conditions a color docs not 
develop, then these ions may automatically be considered to be absent 
in significant quantities, since the test is quite sensitive. By carrying the 
reagent in a water-immiscible solvent like CHC1 3 , one may carry out 
extractions (and hence concentration) of these ions from aqueous solution. 
By comparison to colors developed with solutions of known concentra- 
tion, the method is adaptable to quantitative analysis. The structure of 
the chelate is similar to the one in the preceding example. The concentra- 
tion limit of the test is in the order of 0.002 ppm for Zn+ 2 and 0.05 for Pb f2 . 
See special experiment 10. 

3. Quinalizarin, Carmine, and Benzoin for Boron. The first two 
reagents give color reactions with borates as described in anion test 22-26. 
The solvent used is concentrated H 2 SO 4 . The chelate formed when 
benzoin and borate react in alcoholic solution is fluorescent under ultra- 
violet light. See special experiment 8. All these methods are capable 
of detecting 0.2 ppm of boron. 

4. 8-Hydroxyquinoline (Oxine) for Several +2 and +3 Metals. 
This reagent is well known in quantitative analysis, used for example as a 
precipitating agent for Mg+ 2 and A1+ 3 in light alloy analysis. It also gives 
bulky precipitates under proper conditions with Mn f 2 , Zn f2 , Cd+ 2 , Co 12 , 
Pb+ 2 , Cu+ 2 , Fe f3 , Cr 43 , Ga+ 3 , and some others, about twenty-five in all. 
As in previous discussion, the proximity of electron-donating groups 
makes ring formation by chelation possible. With divalent metals 
(illustrated below) two oxine molecules are coordinated, and with trivalent 



metals, three. This greatly increases the molecular weight of the product 
over the original ionic weight of the metal ion, giving a large volume of 
precipitate from a comparatively small amount of metal tested. See 
special experiments 6 and 8. 

+ M+ 2 = 

M + 2H+ 

This ring system means a carbon atom is at each corner and hydrogen 
atoms are attached wherever needed to complete the structure with four 
valence bonds on each carbon. The molecular formula is C 9 H 7 ON. 

5. Zirconyl Alizarinate for F~. This test in a sense is the opposite of 
the four cases above, in that the disappearance of a colored chelate 
indicates a positive reaction. A chelate formed between zirconium and 
the organic reagent, alizarin, is somewhat weakly constituted and in the 
presence of fluoride is destroyed, accompanied by a color change from 
red to yellow. A stronger zirconium-fluoride complex is formed. By 
visual observation alone, the test will detect 0.1 ppm F~. See anion 
test 22-27 and special experiment 3. 

6. a-Nitroso-$-naphthol for Co+ 2 . A coordination compound is 
formed in HAc solution which has a characteristic red color. The test is 
good in the presence of 100 times as much Ni 1 " 2 as Co +2 , which is valuable 
since these ions occur together in group 3 analysis. The concentration 
limit is about 2 ppm Co+ 2 . See test 17-12. 

7. Sodium Diethyl Dithiocar hamate for Cu+ 2 . In NH 4 OH solu- 
tion this reagent gives yellow to brown colors with cupric ions in the range 
0.02-5 ppm Cu+ 2 , and with more concentrated copper solutions gives 
brown precipitates. The reagent also is used to detect certain other 
metal ions (see references) but is most sensitive for copper. See test 16-8. 

Type B. The reagent and the substance under test are involved in 
oxidation-reduction, the reagent changing in characteristics, usually color. 

Redox is most often associated with bond rearrangement in rings. 
The structural changes are from a benzenoid (reduced form) ring to a 
quinoid (oxidized form) ring and vice versa: 

// ^ oxidation 

X / ) Y ^ * 

\ / reduction 



The presence of electron-donating groups (X and Y) such as OH and 
NH 2 facilitates the reaction. All quinones are colored. 

8. Benzidinejor Certain Oxidizing Agents. A number of oxidants 
like MnO 2 , PbO 2 , NOj, NOj, and some biological substances give this 
test. The reagent is converted from a colorless to a deep blue solution 
(benzidine blue) containing some of the benzidine in the quinoid form. 
About 1 ppm of the oxidants will give the test. See test 22-18 and 
special experiment 7. 

9. Rhodamine-B for Antimony. In the presence of the reagent and 
nitrite ion, Sb m is oxidized to Sb v . The latter then oxidizes the reagent, 
which gives a color change. This test is good to about lOppm of antimony. 
Tin, which occurs with antimony in cation group 2B does not interfere. 
See test 16-12. 

10. Malachite Green for SO~ 2 . As described in test 22-10, this 
reagent is used to detect SO^ 2 in the presence of S 2 Og 2 , although sulfides, 
polysulfides, and some other reducing agents give the same reaction. A 
positive test is the bleaching of the green color as the dye is reduced to its 
colorless (leuco) form with concentrations of sulfite greater than 20 ppm. 

11. The Diazo Reaction for Nitrite. In acid solution, nitrite is in 
the form of a weak acid HNO 2 , which oxidizes aromatic primary amines to 
diazonium salts : ArNH 2 + HNO 2 -> Ar N=N+. The salts are capable 
of reacting (coupling) with other aromatic molecules to produce azo 
compounds whose colors are detectable at low concentration. If sulfanilic 
acid (I) is diazotized (II) and coupled with a-naphthylamine (III), a visible 
color develops with any concentration of nitrite greater than 0.01 ppm. 
The result is synthesis of a red dye (IV). See anions, test 22-5. 

SO a H 


(II) + 

2H 2 


SO a H 



12. Orthotolidine for C/ 2 . A yellow to red color is obtained with the 
reagent and free chlorine. Other strong oxidizing agents interfere. Again 
the colorless form of the reagent is benzenoid and becomes quinoid upon 
oxidation. The method is employed quantitatively to check chlorine in 
swimming pools, disinfecting baths, etc., using permanent color standards 
prepared from color matching K 2 CrO 4 and K 2 Cr 2 O 7 solutions. As little 
as 0.01 ppm C1 2 is detectable. See special experiment 6. 

Type C. The reagent and the substance under test react to give a com- 
pound by direct synthesis without chelation or redox. 

13. Fluorescein for Br~. The yellow dye fluorescein is converted to 
the red tetrabromo derivative, eosin, when a bromide is oxidized to give 
Br 2 and the latter brominates the dye. The test is adaptable to detection 
of this halide in the presence of other halides and thiocyanate, which are 
interferences in most other methods. See test 22-35. 

The concentration limit is about 40 ppm Br -. 



4Br 2 - Br 

Br + 4H+ { 4Br 

14. Resorcinol for C 2 O~ 2 . In dilute H 2 SO 4 , oxalates are in the form 
of oxalic acid and that is reducible with Mg to glycolic (hydroxyacetic) 
acid: 2H 2 + HO C C OH = HO CH 2 C OH + H 2 O. When 

II II \ 

00 O 

resorcinol in concentrated H 2 SO 4 is added, glycolic acid is converted to 
formaldehyde, H C H, and it gradually reacts with the resorcinol and 


air to form a blue colored two-ring quinone. This procedure will indicate 
oxalate at a concentration of about 150 ppm. See anions, test 22-32. 

Type D. The reagent forms a colored adsorption complex ("lake") 
with the inorganic substance being tested. 

Hydrated oxides like A1(OH) 3 have a large surface area and adsorb 
substances strongly. When some organic compounds are adsorbed, they 
change structure and color. These so-called lakes are colloidal at low 
concentrations and appear as colored solutions rather than as colored 
precipitates. The bonding involved is not definitely known, but there 
may be some chemical reaction, as shown in the example below. 


15. Paranitrobenzeneazoresorcinol for Mg+ 2 . With strong bases, 
Mg+ 2 gives Mg(OH) 2 . The surface of this gel adsorbs the reagent, which 
changes color from red to blue. The product given in the equation below 
is hypothetical. The test will detect Mg f2 at a concentration of 10 ppm, 
although similar gelatinous precipitates may also give a positive reaction. 
It may be used as a spot test with a group 4 and 5 mixture. See cation 
test 19-2. 

O 2 N ( ) N=N f J OH + Mg+ 2 + OH- + *Mg(OH) 2 


// ^ 

>>_0_ Mg.CMgCOH),,), + H 2 

' 2 


16. Atuminon (Aurin Tricarboxylic Acid) for AI+*. This test may 
be carried out in slightly acidic solutions where interference due to other 
hydroxide-forming metals is decreased. Hydrous SiO 2 does not give the 
test. The yellowish reagent solution gives a red color upon adsorption 
with Al(OH) ;i . Reaction is probably similar to that pictured above. 
The test will detect 0.3 ppm A1+ 3 although in such dilute solution, 
A1(OH) 3 is invisible. When a large quantity of Al(OH) a is formed in the 
presence of the reagent and the mixture centrifuged, the color of the 
precipitate is bright red. 


1 . (Library) A number of reagents have been described without accompany- 
ing reaction equations. Find reference to these, copy the equations, and in 
each point out the features that classify the reaction according to the four 
divisions given in this chapter. 

2. You are asked to select some organic compounds that might react with 
inorganic ions as described here. Which of the following do you think have 
possibilities? Why? 

(a) CH 3 CH 2 CH, (b) 

(d) HS CH 2 CH 2 NH 2 (e) CH 3 OH (/) CH ;i ~-C ~ CH 2 - C CH 3 

li i! 

o o 

(g) CH 2 F 2 


3. (Library) The symbol R is used in organic chemistry to denote a 
hydrocarbon segment or radical, as CH 3 , C 6 H n , etc. From reference 
works and using R as necessary, show formulas for the general alcohol, ether, 
aldehyde, ketone, carboxylic acid, acid anhydride, amide, amine, nitrite, and 
sulfonic acid. 


1. F. Feigl, J. Chem. Educ., 21, 294 (1944). (Spot reactions) 

2. F. Feigl, J. Chem. Educ., 22, 558 (1945). (Spot reactions) 

3. Organic Reagents Symposium, Anal. Chem., 21, 1298 (1949). '(General) 

4. M. G. Mellon, Anal. Chem., 24, 924 (1952). (Colorimetry) 

5. F. Feigl, Spot Tests, 4th ed., Elsevier, New York (1954). 

6. E. B. Sandell, Colorimetric Determination of Traces of Metals, 2nd ed., Inter- 
science Pub., New York (1950). 

7. F. D. Snell and C. T. Snell, Colorimetric Methods of Analysis, 3rd ed., Van Nostrand, 
New York (1949). 





Metallurgy is a broad field which will be surveyed here under its two 
main subdivisions: process metallurgy, which is concerned with the 
extraction of metals from their natural sources and subsequent refinement, 
and physical metallurgy, which deals with metal combinations, micro 
structures, and prediction of behavior in use. Metallurgy is a very old 
art (viz. 'The Bronze Age"), but a newcomer among the sciences. Since 
a good portion of the qualitative analysis course is concerned with metal 
analysis, and because about 80 of the 101 elements now known are metals, 
a brief survey of metals, minerals, and alloys is needed to increase the 
analyst's viewpoint and understanding of his own specialty. 

Sources of Metals 

Metals are obtained from two sources: the earth's crust and the ocean. 
Although only Mg is commercially extracted from the latter, its potentialities 
are being further explored as the earth's supply is rapidly dwindling and 
as atomic power promises to give cheap energy for the evaporation of sea 
water. In examining Table 13-1 it must be kept in mind that the figures 
are averages and that only concentrated ore deposits, which are relatively 
rare, are economically feasible for exploitation. Thus, titanium and 
aulminium are spectrographically found in almost all rocks and clays, 
but recovering them from low-grade ores is at this time only a laboratory 





(a) Elements In The Earth's Crust 


Per Cent by 


Per Cent by 


Per Cent by 



































Other rare 




















All others 


(/>) Elements In Sea Water 


Parts per 


Parts per 


Parts per 














1 272 

















































0.2-3 x 10- 10 



Minerals and Ores 

Minerals or ores are naturally occurring free metals or chemical com- 
binations of metals with other elements. Mixtures of minerals are called 


rocks. Ores are found in a few places in the earth in sufficient concentra- 
tion to warrant extraction. The mineral is first mined, commonly by 
digging or washing out with water. In most cases it is not of sufficient 
purity to be sent directly to chemical processing, so it first must undergo 
ore dressing, a term applied to any preliminary purification. This rids 
the useful portion of surrounding matrix or gangue, and the concentration 
process is known as ore beneficialion. These methods must be cheap, 
rapid, and simple because of the large tonnages to be handled daily. 
Matrix material finds use as gravel in road building, fills, etc. 

Beneficiation Methods 

(a) Magnetic separations are used for some iron and other magnetic 
ores. The rock is crushed so that a minimum of matrix is carried along. 

(b) Water or acid leaching processes may be used to selectively dissolve 
either ore or gangue. Low-grade copper (oxide) ore, for example, is 
handled in this way; H 2 SO 4 dissolves the oxide, giving a cupric sulfate 

(c) Specific gravity separations are conducted on a small scale by the 
gold panner, or on a large scale in tanks where crushed minerals are 
separated from matrix by floating one or the other ofT and allowing the 
rest to settle. The flotation process is a modification of this in which 
finely crushed ore called pulp is mixed with water, soap, an oil, and air. 
The froth produced preferentially wets certain ore components, usually 
the more dense mineral itself and floats it off, leaving the worthless matrix 
behind. The recovery of the process is good and tailings left from previous 
less efficient beneficiation processes have been reworked profitably. 

(d) Other chemical separations besides acid leaching are used in some 
special cases. One example of this is the Dow process for winning 
magnesium from sea water. Calcined oyster shells (CaO) are added to 
the water, precipitating Mg(OH) 2 . This is filtered off, dissolved in HC1, 
and evaporated giving MgCl 2 H 2 O. A great concentration of magnesium 
has obviously been effected. 


After beneficiation, the dressed ore is ready for reduction to the free 
metal if it does not occur native (free) as an original mineral. At least 
one chemical reaction is necessary. 

(0) Pyrometallurgy is characterized by being a high temperature process. 
If the metal is sufficiently low in the activity series to be native, it is simply 
melted away from the less fusible gangue. If the metal oxide is the 
mineral, or the metal has been roasted previously to the oxide, carbon 
(coke) is the usual reducing agent. If easier, the gangue may be made 


fusible (lower melting) by adding a flux to it, and the metal is separated 
from this liquid at high temperature. Matrixes are often silicaceous 
(SiO 2 ) or limestone (CaCO 3 ) types. At high temperature, silica is acidic 
and CaO (from calcium carbonate decomposition) is basic, and they 
neutralize each other to form fusible calcium silicate glass slags. Thus if 
one wishes to fuse a silica matrix like quartz, he uses a basic flux like lime- 
stone, and vice versa. In the furnace both metal reduction and slag forma- 
tion take place with the slag rising to the top and the molten metal being 
tapped off the bottom. The following is typical : 

C + Fe 2 3 + Si0 2 + CaC0 3 4> CaSiO 3 + CO + CO 2 + Fe 

Reducing mineral matrix flux Molten gases molten 

agent slag metal 

Furnace linings must be able to withstand the temperatures and reactions, 
and they are gradually dissolved as slag although refractory (high melting) 
oxides like A1 2 O 3 and Cr 2 O 3 are used. The role of carbon in these reactions 
is most important since it is the principal reducing agent: 

C + CO 2 = 2CO 

Carbon monoxide can also work as a reductant but excess carbon is needed 
to prevent CO 2 from reoxidizing metal: 

C0 2 + Fe = FeO + CO 

(b) Hydrometallurgy processes are run at ordinary temperatures in 
aqueous solution. Minerals containing more noble metals are dissolved 
in acid or with complexing compounds and the metals are displaced by 
adding cheaper, more active metals. For example, NaCN dissolves 
Ag and Au as cyano complexes which arc decomposed when powdered 
zinc is added : 

2Ag(CN)~ + Zn = 2Ag + Zn(CN 2 

Another example is one already mentioned wherein copper oxide, while 
still in ore pockets, is leached from an inert matrix with acid. The acid 
is allowed to react underground for some time, then pumped up to tanks 
where scrap iron ("tin" cans) is added to displace copper. Tons of cans 
are collected daily in large cities and magnetically separated from other 
refuse for this purpose. 

(c) Electrometallurgy is the process in which metals are electroplated 
from aqueous solutions at ordinary temperature or from anhydrous molten 
salt solutions at high temperature. An example of the latter is the 
electrolysis of magnesium chloride in the final step of the Dow process to 
produce magnesium metal and chlorine. Similar methods to obtain 
aluminum (Hall process) and alkali metals (Downs cell) are described with 
the chemistry of those elements in later chapters. 


(d) Aluminothermy or the Goldschmidt process is used on oxides which 
for one reason or another (as carbide formation) are not satisfactorily 
reduced with carbon. The powdered oxide is mixed with fine aluminum 
particles and heated. Enough heat is liberated as A1 2 O 3 forms to give 
molten metal as the main product : 

Cr 2 3 + 2A1 = A1 2 3 + 2Cr 

(e) Special methods are described for some of the refractory metals in 
Chapter 20. Because of their high melting points, ordinary procedures for 
metal recovery and general metallurgy do not work. Their high tempera- 
ture applications have nevertheless centered a great weight of recent 
metallurgical research on metals like Be, Ti, Zr, Hf, V, Cb, Ta, Cr, Mo, 
and W. The radioactive metals of the actinide series and other active 
wastes of the atomic energy operations also present major difficulties in 
handling because of the radiation and newness of the field. For example, 
the Hanford works for plutonium separation were built on the basis of 
micro experimentation with a mass of Pu practically invisible to the naked 
eye a scale-up of an unprecedented 10 10 times. 


When the metals have been obtained in semipure form by one of the 
above processes, they may find use as such or be further refined. 

(1) Electrolytic refining is as described Imder electrometallurgy, p. 202, 
although here a semipure metal is used as the anode, a pure bar or plate 
of the same is the cathode, and a solution of one of its salts is the electrolyte. 
In this way copper of purity better than 99.9 % is produced. For very high 
melting metals like Ti, W, etc., sintered bars are melted by electrical 
resistance in special crucibles under vacuum or in an inert atmosphere 
(see Chapter 20). 

(2) Distillation and sublimation are used on metals such as Hg, Zn, and 
As, which are volatile. The vapor is condensed and may be redistilled or 
resublimed to give products almost 100% pure. 

(3) Smelting at high temperatures is needed to purify many metals. 
Fluxes and degasifying materials are added to lower the melting tempera- 
ture, float off impurities, and keep gases from dissolving, which might 
otherwise embrittle the metal. Aluminum, for example, is added to 
remove O 2 and give a fine grain structure in steel. 

Mineral Types and Their General Treatment 

Which methods of metallurgy one uses will be determined by all the 
factors involved, from ore type to availability of flux materials and energy 
for reductions. Of the common mineral types, five are worthy of mention : 


(a) Oxide minerals are usually reduced with coke (e.g., Fe 2 O 3 , CuO). 
Some noncarbon-reducible oxides are those of Ca, Sr, Ba, and Al (Table 
1-1), and in addition, Cr, Mn, and some other transition metal oxides 
cannot be treated well in this way because they dissolve carbon and form 
very hard, brittle carbides. Their oxides may be reduced by aluminothermy 
or special methods. 

(b) Volatile metals are distilled. As an example, in the Pidgeon process 
for magnesium, finely divided ferrosilicon is pressed into briquettes with 
roasted dolomitic limestone and heated in evacuated retorts at 1150C. 
The magnesium collects in large crystalline masses at the cool end of the 

FeSi + CaO + MgO A > Fe 2 O 3 + CaSiO 3 + Mg 

(c) Carbonate minerals are calcined (roasted) and treated as above: 

MgCO 3 = MgO + C O 2 

(d) Sulfide minerals are roasted in air to convert them to the oxides 
whose handling is described above; SO 2 as a by-product can be converted 
to SO 3 and H 2 SO 4 . Industrial processes are typified by utilization of 
everything of value in the operation. 

(e) Miscellaneous minerals handling includes electrolysis and hydro- 
metallurgy as previously described. 

The relatively small number of important mineral types is a consequence 
of the comparative abundance of the elements and their chemical combina- 
tions at high temperature when the earth was in process of formation. 
Seven types cover most of the cases : 

(a) Native (free): Ag, Au, Pt, Pd, Cu, Hg, Bi, Sb, As. As expected, 
the easily reduced metals, low in the replacement series (Table 1-1), are 
the ones forming unstable and few compounds. 

(b) Oxides: Fe, Sn, Mn, Al, Cu, Zn, Ti, U, Zr. The same metals 
will of course burn in air to give these oxides. 

(c) Sulfides: Fe, Cu, Zn, Ni, Co, Sb, Pb, Cd, Hg, Mo, V. Familiar 
metals are from group 2 and 3 cations. 

(d) Sulfates: Pb, Ba, Ca, Sr. The same sulfates are used to separate 
and identify these metals in qualitative tests. 

(e) Carbonates: Ca, Mg, Sr, Ba, Fe, Cu, Zn, Pb. The first four may be 
precipitated with (NH 4 ) 2 CO 3 in cation group 4. The others form sulfides, 
in previous separations, that are more insoluble than their carbonates. 

(/) Halides: Na, K, Mg, Ag. The first three occur in natural brines 
and the ocean. 
(g) Silicates: K, Li, Al, Be. The metals are not reasily recovered. 


Metal Characteristics 

No sharp dividing line separates metals from nonmetals, and borderline 
cases will have some chemical and physical attributes of both types of 
element. It is helpful, however, to have in mind the most obvious 
differences before further elaboration on the metallic state. 

(a) Physical Properties of Metals. 1. High luster. Metals reflect 
visible light well although reflectance varies considerably with the wave- 
length of incident radiation. In a finely divided state many metals appear 
black, but in massive form they are silvery, with the exception of Cu and 
Au. Use is made of metallic reflectivity in mirrors and reflectors. Silver 
reflects 95 % of the visible light falling on it, platinum 65 %, and copper 
(one of the poorest) only 55%. 

2. High electric conductivity. The conductivity varies with different 
metals and for a given metal is inversely proportional to temperature. 
Bismuth and mercury are not especially good electric conductors; silver, 
the best, is about sixty times better than mercury. Metals, with their low 
electrical resistance, are used in wires, bus bars, etc., where good conduction 
is needed, while nonmetals, with high resistances, are used as insulators. 
Electric conductivity involves electron movement without metallic decom- 
position. Some metals like Hg, Sn, and Pb exhibit super electrical con- 
ducitivity at temperatures in the range (MO A, current flowing for long 
periods after a potential across them is removed, due to exceptionally low 
electrical resistance. 

3. High heat conductivity* The order is roughly parallel to that of 
electrical conductivity trends. The generally high melting points and good 
heat conductivities make metals useful in all types of heat exchange 
equipment from cooking pots to condensers, radiators, and boilers. 
Nonmetals are insulators. 

4. High density. Compared with other substances, metals are usually 
more dense. Some exceptions occur with the lighter metals, but, lower 
in the periodic table where metallic properties are pronounced, densities 
are high. Alloys containing large amounts of dense elements have high 
densities also, of course. 

5. High strength. Mechanical strength of materials is measured and 
expressed in various ways. The ultimate tensile strength, for example, is 

* The relation between heat and electric conductivities has been studied. The Law 
of Wiedemann and Franz gives a simple relationship which approximately describes 
many cases : 

Thermal conduct, x 10 8 /Electncal conduct. KT 

where T is the absolute temperature and K is, approximately a constant. 


defined as the axial load per unit of cross-sectional area (pounds per 
square inch, psi) needed to rupture a prismatic test specimen in a machine 
designed for the purpose. Other measures of strength express charac- 
teristics of elasticity, creep, crack resistance, etc. 

6. High hardness. Hardness is defined as resistance to permanent 
deformation by indentation, as measured on standard machines which 
indent the metal surface with a diamond or hard-alloy stylus under given 
loading. The test results are expressed in arbitrary units deriving their 
names from the machines used: Brinnel, Rockwell, Vickers, etc. (For 
geologist's hardness measure, find library reference to Mohs' scale.) 
Although some light nonmetallic elements and their compounds are hard, 
such as diamond, silicon carbide, and boron nitride, the metals are more 
generally typified by high hardness. 

7. High deformability. This is manifested in two ways: ability to be 
hammered into thin sheets (malleability), and ability to be drawn into fine 
wires (ductility). A few metals like Pb, Bi, and Sb are poor in this respect 
but most are superior to nonmetals. Metals are thus said to be capable of 
plastic flow with large elastic limits. The combination of high yield strength 
and great capacity for plastic deformation is the most important mechanical 
characteristic of metals. 

8. High melting and boiling points. There are nonmetallic materials like 
graphite and silicon carbide that are refractory; that is, they retain their 
shape and original composition at high temperatures. However, it is 
more characteristic of metals than nonmetals to have high melting and 
boiling points, but even the melting and boiling points of metals are 
scattered on the temperature scale between the extremes of mercury and 
tungsten. High temperature applications of metals as in jet aircraft have 
brought new demands on metallurgists to develop alloys with refractory 
properties, and attention is automatically directed to the transition metals 
that have low creep at high temperature. 

9. Crystalline. All metals are typified by being crystalline, and most of 
their space lattices are simple (see the section on metal structures, p. 211). 
Alloys arc crystalline also and may have simple or complex lattices. 

(b) Chemical Properties of Metals. 1. Metals usually form the 
cation of a chemical compound and during formation from the elements 
became positive ions by electron loss and act as reducing agents. Metals 
appear not infrequently in the anion also, as Zn(OH)^ 2 , AsS^~, Fe(CN)^ 4 ^ 

2. Metals form oxides that give basic reactions. Thus, dissolving 
calcium oxide in water gives the base calcium hydroxide or dissolving it 
in acid gives a salt. Exceptions to this are amphoteric oxides like ZnO 
which dissolve in either acid or base, showing that they can act as weak 


bases or acids, and some higher valent oxides like Mn 2 O 7 , which give 
weakly acidic reactions. 

3. Metals replace hydrogen in acids. See Table 1-1. 

4. Metals have low electronegativity values although a few nonmetals 
like H also have a strong tendency to form positive ions. See Fig. 3-14. 

The above properties are best demonstrated by elements in the lower left 
portion of the periodic table. 

Metallic Valence 

The binding forces in metals interest engineers as well as chemists, and 
explanation of metallic bonds has contributed to general understanding in 
both their fields. 

The first theory of the metallic bond is due primarily to the work of 
Drude, Losenty, and Lorentz at the beginning of this century. They 
proposed that metal atoms be thought of as spheres vibrating about points 
in a crystal lattice, and surrounded by clouds of moving electrons belonging 
to no particular atoms. This condition is brought about by each atom 
tending to give electrons because of the electropositive character of metals. 
The metal structure is one of positive centers arranged in regular geo- 
metric pattern through the interstices of which move free electrons, the 
whole bonded together by attraction of the oppositely charged bodies. 
The theory was successful in explaining many metallic properties. Thus, 
high electric and heat conductance is due to readily mobile, energy- 
carrying electrons, and opacity is due to loosely held electrons of various 
energies absorbing and re-emitting light of all visible wavelengths. Since 
atomic vibration is directly proportional to temperature, it is expected that 
high temperature will hinder electron movement because vibrating atomic 
centers will scatter the electron wave front. Electric conductance in 
metals is therefore inversely proportional to temperature. Electrical 
resistance approaches zero and conductivity approaches infinity near 
absolute zero. Alloys as cast have a disordered atomic structure and the 
conductivity is usually lower than that of the pure metals. Heat treat- 
ment will allow the atoms to move into a more ordered pattern with better 
conductivity and some heat treatment results are followed by such 

Nonmetals generally are oppositely constituted. In substances like 
SiO 2 , electrons are tightly held in formal covalent bonds, and, not being 
capable of movement, are not available to conduct heat or light or to 
intercept radiations; they are therefore insulators and are transparent. 
There are also some borderline substances called semiconductors which 
conduct under certain conditions. These and nonconductors become 
conducting if energy put into them is sufficient to break bonds and produce 


unattached electrons. In this way silicon and germanium crystals become 
conducting upon heating to moderate temperatures, and even diamond is a 
conductor under high energy ray bombardment. This is the principle 
behind "crystal counters" devices used to detect and measure radiation. 
Electrons thus loosened move under applied voltage and constitute an 
electrical current. The bond robbed of an electron now is a positive 
"hole" in the structure and electrons will drift toward it, creating holes 
where they depart and giving the effect not only of moving electrons but of 
oppositely moving (positive) holes which also contribute to conductance. 
These ideas have been refined by application of the quantum theory by 
Fermi, Sommerfeld, and Dirac, but are beyond the scope of this text. 

Metal atoms are capable of forming many types of bonds, and electrons 
resonate among various positions. This not only helps explain stability 
and conductivity but also ductility and malleability. 

When a metal ingot is rolled into a sheet, bonds are broken as atoms are 
moved away from each other, and this can result in fracturing a crystalline 
substance. Metals, however, are not dependent on single sets of bond 
types, distances, and angles, and reformation of bonds among new 
neighboring atoms gives a sheet about as well knit electronically as the 
original ingot. Metals toward the bottom of the periodic table are best 
capable of deformation, since they have only small energy differences 
between possible bonding levels and bonds break and form with little 
change in total energy. 

For bonding, each atom must have unpaired electrons to donate to the 
bonds and empty orbitals in which to accomodate accepted electrons. The 
maximum number of covalent bonds is limited in elements on the left 
side of the periodic table by too few electrons, and on the right by 
too few empty orbitals. Thus it is that maximum covalence occurs in 
elements in the middle of the periodic table with a resultant maximum 


Ferromagnetism is related to electron structure. It is a property of Fe, 
Co, and Ni, the three elements which are strongly drawn into magnetic 
fields and exhibit lasting magnetism after being withdrawn. It is postu- 
lated that in these metals there exist domains of magnetism which feature 
groups of atoms showing electronic displacement in single directions. 
Under magnetic field influence, the domains themselves align, and their 
magnetic effects reinforce each other. If the metal is then withdrawn 
from the field, it remains magnetized until heated or severely deformed, 
whereupon the domains return to a random orientation and magnetism 
is destroyed. 


Conditions for domain existence appear to include both unfilled orbitals 
and a particular internuclear distance, and the three elements mentioned 
above are the only one having both. From its position in the periodic 
table, electron arrangement, and atomic radius, one suspects manganese 
might be ferromagnetic, but apparently the nucleii are a little too close to- 
gether. It is interesting to note that certain Mn alloys containing non- 
magnetic elements such as Cu, are ferromagnetic, presumably because now 
the Mn atoms are displaced to the proper distances. 

Crystal Chemistry 

Obviously if one knew the arrangement of atoms in a metallic crystal 
he would be in an excellent position to understand many chemical and 
physical properties of all metals and crystalline materials. It is also 
evident, however, that because of the minuteness of individual atoms, 
ordinary methods of measurement will be to no avail. 

The discovery of X-rays occurred 17 years before M. von Laue, in 1912, 
conceived of a use for them in crystal measurements. It was known that 
parallel lines (a diffraction grating) ruled on a surface would scatter (diffract) 
visible light incident upon them if the distance between lines was in the 
range of the wavelengths of light. It was Laue's idea that if one assumed 
crystals to be built up of parallel layers of atoms as close together as 
X-ray wavelengths, perhaps such planes within crystals might act as 
diffraction gratings for X-rays. The experiment was tried by Friedrich 
and Knipping, who exposed a thin, cubical crystal of ZnS to X-rays. They 
obtained a photograph showing spots in an arrangement having cubical 
symmetry. Since their X-ray tube simultaneously produced X-rays of 
many wavelengths, all crystal planes of high atomic population, although 
various distances apart, were shown present by having diffracted radiation 
of some wavelength. Mathematical translation of the photo pattern is 
capable of giving the angles between planes and hence the shape of the 
structural unit, though not its dimensions. This unit cell is visualized as 
containing enough atoms to describe its symmetry, and, if repeated in 
three dimensions many times, will yield the visible crystal. The crystal's 
external form thus depends upon internal atomic arrangement. 

The following year, W. H. Bragg and his son, W. L. Bragg, began 
experiments to find whether or not crystal planes might reflect X-rays. 
Since X-rays are not significantly bent upon passing through matter (as is 
visible light), they should penetrate and reflect without distortion, thus 
making accurate measurements on reflectance angles within crystals 
possible if monochromatic rays are used. This was proved by the 
Braggs, and from the geometry of the setup they showed that know- 
ledge of these angles allows calculation of interatomic distances. The 



relationship is stated by the Bragg equation (derived in many physics 

nX = 2d sin 6 (13-1) 

Theta is the angle of incidence and reflection of monochromatic X-rays 
having wavelength L As is increased from C, a progression of angles 
is found at which reflection reinforcement from successive parallel sheets 
of atoms takes place. These positions are called the order of reflection 
and correspond to n = 1, 2, 3 etc. At other angles reflection inter- 
ference takes place because reflected rays are out of phase. Term d is 
the distance between planes. Since each crystalline material has its own 
atomic arrangement and atomic dimensions, each gives a different series 
of 6 values for which the Bragg equation is satisfied. These may be 
compared with values obtained by measurements on known substances, 
and unknowns can be identified. 

FIG. 13-1. A recording X-ray spectrometer. As the crystal is rotated, 6 changes, 
reflection reaches the counter, and the pen traces a graph characteristic of the sample's 

internal arrangement. 

Structure determination from Bragg data begins by computing inter- 
planar distance ratios once d values are found. The ratios are then 
compared to ratios calculated for probable geometrical arrangements of 
atoms until, by trial and error (or more recently by automatic punch card 
machines which sort out data), a fit is obtained. 

A simple two-dimensional analogy is as follows. Suppose someone 
has arranged many marbles on a table in a regular repeating pattern and 
covered it with a sheet of cardboard. You are then asked to find what the 
arrangement is, based only on whatever measurements you can make as 
you squint down the rows of marbles from various positions around the 
edge of the table. By means of a ruler, interrow distances can be found 



and ratios of these distances can be calculated by choosing one row as a 
reference. The ratios are then compared to those calculated by geometry 
for different marble arrangements drawn on paper. The pattern that fits 
the ratios is assumed to be the correct one. When this process is applied 
to a three-dimensional pattern containing several different sizes of atoms, 
it is considerably more complicated. 

Metal Structures 

Of the many possible geometrical patterns metal atoms might assume 
upon crystallization from a melt, only three are common. These are 
high-symmetry types where maximum coordination is possible; i.e., a 
given atom is within bonding distance of a maximum number of other 
atoms. Such structures are expected to make bond resonance easy, 
stability high, and be dense and hard; all general characteristics of metals. 

If the atoms are imagined as spheres, there are two types of closest 
packing of spheres: cubic closest and hexagonal closest packing (both of 
which have 12-coordination), and a third structure called body centered, in 
which a given atom has eight near neighbors and six others only a little 
more distant, giving 14-coordination. X-ray analysis of metals shows 
these three space lattices are the usual ones. 

1. Cubic closest packing. This may be thought of as an expanded 
simple cubic lattice with six additional atoms one in each face of the 
cube. For this reason it is called a face-centered cubic lattice. As seen 
in Fig. 13-2, the unit cell, each atom may form 12 bonds of about equal 
length. Metals crystallizing in this pattern are Ca, Sr, Al, Sc, La, Ce, Th, 
U, Fe, Co, Rh, In, Ni, Pd, Pt, Cu, Ag, Au, Tl, and Pb. 

FIG. 13-2. Two views of cubic closest packing which gives the face centered cubical 
lattice. Twelve-coordination is demonstrated in the right-hand figure. 

2. Hexagonal Closest Packing. The unit cell plus four atoms to 
illustrate 12-coordination is shown in Fig. 13-3. It is to be noted that 
atomic population is greater along the basal planes than in other directions. 



Compressibility is least and electric conductance greatest in these planes. 
Slippage of sheets of atoms may occur under stress along the same planes 
but is resisted in other directions. The cubic lattice above, on the other 
hand, has four equivalent planes of high atomic population, and, due to 
this, metals of the close packed cubical structure are usually most ductile 
and malleable. Metals having the hexagonal lattice are Be, Mg, V, Ti, 
Hf, Re, Ru, Os, Zn, Cd, Pr, Nd, and Er. 

FIG. 13-3. Two views of hexagonal closest packing showing 12-coordination. Com- 
pare these to Fig. 132. Note alternate anangement of planes. 

3. Body -centered cubic. This lattice is not as tightly packed as the 
other two. A given atom has eight atoms equidistant from it and six 

FIG. 13-4. The body-centered cubic lattice. The right-hand figure demonstrates the 
14-coordination. Body-centered atoms are shaded darkly for contrast. 

others 15% further removed with which it may also form bonds, so the 
coordination number is said to be 14. Selecting the atom in the center 
for reference, its eight nearest neighbors are at the corners of the cubic 


lattice and the six others are the body-centered atoms of the six adjacent 
cubes. Metals crystallizing in this form are Li, Na, K, Rb, Cs, Zr, V, 
Cb, Ta, Cr, Mo, W, U, and Fe. 

It will be noted that several metals are capable of crystallizing in more 
than one of these three common lattices and the polymorphism (several 
forms) of some include rarer lattices as well. 


1 . Why are minerals commonly the oxides, sulfides, carbonates, and silicates 
of metals rather than say, acetates, nitrates, and perchlorates ? 

2. J. P. Slipshod reasons that increasing temperature should increase the 
kinetic energy of electrons and so make metals better electric conductors at 
high temperatures than at low temperatures. J.P.'s Uncle Frisbee argues, on 
the other hand, that electrons are too small and bound too tightly to be affected 
by heat, so electric conductance should be temperature independent. Comment 
on these opinions. 

3. (Library) From data to supplement tables in this book, plot atomic 
numbers (abscissa) vs. melting points of the elements. Repeat for numbers vs. 
densities. In a few sentences summarize the trends. List the extreme values 
from both compilations to emphasize the spread. 

4. (Library) For each of these metals, give briefly the main minerals, metal- 
lurgy, alloys, properties of the metals, valence states, and commercial uses of a 
few important compounds and alloys: (a) Ce (b) Pd (c) Ga (d) Ge (e) Th. 

5. Early work on crystals was done by E. Mitcherlich, who announced in 
1819 his rule of isomorphism. This states that compounds of similar chemical 
formula, such as KH 2 PO 4 -H 2 O and KH 2 AsO 4 -H 2 O, often form crystals with 
the same external form. Show how the concept clarifies this problem: metal 
jc -forms a sulfate isomorphous with zinc sulfate heptahydrate, containing 
19.81 % x. Find the atomic wt of x. (Ans. 54.9). 

6. Consider a simple unit cube of a metallic element comprising eight atoms. 
If that cube is completely surrounded and its atoms shared by bonding to similar 
structures on all sides (a) how many atoms may be said to exclusively belong to the 
reference cube? (h) If the side of the cube is 3.40 A and the density is 8.00 g/cc, 
what is the element's atomic wt? (Ans. (b) 188) 

7. Certain planes of atoms in a crystal are known to be 2.80 X 10 ~ 8 cm 
apart. Second-order X-ray reflection is found to occur at an angle of 12 50'. 
Show that one can find the wavelength of the X-rays. 

8. (Library) (a) Copy the structures of diamond and graphite and describe 
how these explain the radically different properties of the two allotropic forms 
of carbon. 

The lubricating ability of layer lattice materials like graphite and MoS 2 has 
been attributed to slippage of sheets of atoms, made possible by relatively long, 


and hence weak, bonds between sheets. More recently it has been postulated 
that air molecules adsorbed between the sheets weaken the intersheet bonds and 
the sheets slide on these air "cushions." In one experiment, ordinary graphite 
was heated in an evacuated chamber, then tested as a lubricant. It was found 
to be abrasive. Explain how the experiment bolsters one of the above theories. 
(b) It is known that when potassium metal is heated with graphite, the metal 
is absorbed, forming a red compound, C 8 K, and a dark-blue compound, C 16 K. 
X-ray measurement shows that the distance between carbons in the layers is the 
same as before but the distance between layers has increased. Postulate 
structures with rough sketches. 

9. (Library) Find reference to (a) electron diffraction (b) neutron diffraction 
(r) powder method of P. J. W. Debye and P. Scherrer, and (d) goniometer of 
W. H. Wollaslon. Sketch the device inferred in each and give brief qualitative 
statements concerning operation and use. 

10. From Table 13-1 (b) Calculate the number of pounds of gold in a cubic 
mile of sea water. (D = 63.86 lb./ft 3 ). How many tons of Mg could be 
recovered in that volume of water? 

11. In metallurgical processes, explain why sodium carbonate is a basic flux 
and borax is an acidic flux, yet both give basic reactions in water. Give specific 

12. The following are high-temperature metallurgical reactions. Give the 
balanced equations. 

(a) Carbon monoxide reacts with antimonous oxide, producing antimony 
metal and carbon dioxide. 

(b) Silica reacts with calcium orthophosphate and coke to give calcium 
metasilicate, phosphorous, and carbon monoxide. 

(c) Ferrous metasilicate is fused with calcium carbonate, producing iron(ll) 
oxide, calcium metasilicate, and carbon dioxide. 

(d) Boric oxide and cuprous oxide are heated with air in an oxidizing flame 
and copper(ll) metaborate is the product. 

(e) Potassium carbonate, bismuthous sulfide, and oxygen react to form 
potassium bismuthate, potassium sulfate, and carbon dioxide. 

(/) Manganese dioxide plus potassium hydroxide yields potassium manganate. 

(g) Ferrous chromite, oxygen, and sodium carbonate fused together give 
iron(III) oxide, sodium chromate, and carbon dioxide. 

(/*) Manganese heptoxide fused with plumbous oxide yields lead(II) 

(/) Fusion of lead(II) sulfide, potassium nitrate, and calcium oxide produces 
a mixture of calcium sulfate, potassium nitrite, and calcium plumbate. 

(j) Sodium metaphosphate and manganous oxide give a single product, 
sodium manganese(II) orthophosphate, upon fusion. 

13. A certain compound of elements X and Y crystallizes in the body-centered 
cubic lattice. X atoms occupy the body-centered positions while Y atoms form 
the corners of the cube. J. Slipshod reasons that the simplest formula must be 
XY 8 , but the sharpies in class conclude it is XY. Explain. 

14. Consider three separate structures consisting of rigid equivalent spheres 


2 Angstrom units in diameter and representing atoms in cubic symmetry. Con- 
sider further that imaginary lines connect centers of spheres at the corners of 
the cubes. Structure A is simple cubic, B is body-centered, and C is face- 
centered, (a) Add up the whole spheres and fractions of spheres contained 
within the cubes described by the imaginary lines for each structure. (/>) Find 
the volumes of the cubes within the imaginary line boundaries, (c) Find the 
volumes of total sphere material within each cube, (tl) By comparing the last 
two answers, determine very approximately what fraction of each cube is 
occupied by spheres. (Ans. (tl) A, J B, f C, J) 


1. F. C. Brenner, J. Chem. Educ., 25, 371 (1948). (Crystal models) 

2. H. K. Work, /. Chem. Educ., 28, 364 (1951). (Metalluigy) 

3. R. A. Lefever, J. Chem. Educ., 30, 554 and 486 ( 1953). (Metallic conductance) 

4. V. N. Ipatieff, J. Chem. Educ., 30, 1 10 (1953). (Metal displacement) 

5. I. Asimov, J. Chem. Educ., 31, 70 (1954). (Earth's crust) 

6. D. W. Sawyer and R. B. Mears, Anal. Chem., 17, J (1945). (Metal testing) 

7. H. F. Beeghly, Anal. Chem., 22, 235 (1950); 23, 228 (1951); 24, 252 (1952). 
(Ferrous metallurgy) 

8. M. L. Moss, Anal. Chem., 24, 258 (1952). (Nonferrous metallurgy) 

There are many excellent periodicals, texts, and reference books on general and 
specific metallurgy. Publications from The American Society for Metals are especially 
recommended. In addition, trade literature from the large metal companies give the 
very recent developments and industrial applications. These include organizations 
like U.S. Steel, Allegheny-Ludlum Steel, Aluminum Company of America, American 
Brass, Westinghouse, General Electric, Bridgeport Brass, Chase Brass, Bell Telephone, 
Reynolds Metals, International Nickel, Crucible Steel, Bethlehem Steel, Haynes 
Stellite, Jones and Laughlin Steel, Norton, Republic Steel, Revere Copper and Brass, 
Ryerson and Son, etc. References to specific metals are given at the ends of Chapters 
15, 16, 17, 18, 19, and 20. 





Semimicro laboratory methods are less expensive than macro ones as 
well as being more rapid, but, because of the small quantities of materials 
handled, good technique is mandatory. For example, a speck of rust 
from the ring stand or wire gauze may pass as an unnoticed contamination 
if one is working with a 20-ml sample, but it gives a strong test for iron if 
the sample volume is only 10 drops. It is incumbent upon the good 
student, therefore, to know and follow the procedures carefully and to 
organize his work neatly with clean equipment and pure reagents. By 
contrast, the sloppy, impatient operator who "borrows" neighbors' 
chemicals, seldom cleans any glassware, makes illegible notes on scraps of 
paper, and looks at the lab directions for the first time as he runs the 
unknown sample can expect only uncertain analytical results. 


The suggested standard complement of locker equipment is listed in 
appendix Tables A4-8. The following items, if not present, should be 
made, as their use is described in later procedures. 

1. Four Dropping Pipets (Syringes) and Calibrations. These 
pipets are in addition to two (machine made) droppers supplied with the 
regular apparatus. Two may be made from 6-mm tubing and two from 
8-mm tubing. Steps in this preparation are illustrated below. A piece 
of tubing is cut about 8 in. long and heated in the middle by holding the 
ends and rotating it in the flame of a burner equipped with a wing top. 




When pliable, the tube is held vertically and stretched (1) as the operator 
sights down through the tube to keep it straight. When cool, it is cut in 
two at some convenient point in the narrow section and the tips fire 
polished (2) to give fine openings. A lip is turned on the big end (3) by 
heating it and manipulating the glass with the end of a file or a sharpened 
carbon rod (as salvaged from small batteries). The lip may be reheated 
to plastic workability and turned over more by rotating it oh a charcoal 
block so that its shape is convenient to hold a rubber bulb. Finally the 
pipet is calibrated, with the aid of a small cylinder, to deliver some com- 
monly used volume of water like 1.0 or 0.5 ml and a label is pasted on or 
a file mark made to indicate the calibration (4). All labels meant to be 
permanent must be sealed with a coat or two of label varnish or paraffin 
(dissolved in benzene). The final length of the glass portion should be 
about 2.5 4.5 in. 

FIG. 14-1. Operations in ma king droppers. 

For ordinary machine-made droppers, tips are fairly uniform at 3 mm 
O.D. and 2mm I.D., and these droppers can be used most handily for 
dropwise measurements. The cautious student may wish to calibrate 
one or two machine droppers, as time allows, or the following table can 
be used. Tip diameter and surface tension, density and temperature of the < 
liquid are the factors governing the size of drops delivered. 

2. Three Stirring Rods. These may be made from 4-mm tubing by 
melting both ends shut or from 4-mm rod as described below. The rods 
may have different end shapes for special jobs and should be measured for 
length using the containers in which they will be employed. One rod will 





H ? 

Dilute H 2 O solutions 
Concentrated HCl 
Concentrated HNO 3 
Concentrated H 2 SO 4 
Glacial HAc 
6 M NaOH or NH 4 OH 
Concentrated NH 4 OH 

Drops per Milliliter 



Milliliters per Drop 


be needed with a blunt end for crushing solids in spot plate holes (1), 
another may have a cane shape for stirring and resting in 4-in. test tubes (2), 
and a third may be made to fit the 20-ml beakers (3). The only thing to 
remember is to make them long enough so they can be used in hot and 
corrosive solutions without danger to the ringers, yet not so long that they 
will make the containers top heavy and subject to tipping. 

FIG. 14-2. Stirring rods with different shapes. 

3. Student Reagent Bottles. Bottles in the student locker which con- 
tain the chemical supply must be labeled legibly and completely, and a 
protective lacquer used over the paper. Before and after filling from stock 
bottles, all labels should be checked and of course droppers should never 
be interchanged. Dropping-bottle droppers are fairly uniform and cali- 
bration of one or two will correlate drop and milliliter measurements. 
Concentrated NH 4 OH should not be stored in the locker with concentrated 
acids as everything sqon becomes coated with ammonium salts. The same 
applies to NH 4 OH solutions unless this is occasionally unavoidable, and 
such solutions are to be well stoppered. Acids, including acid-containing 


reagents and H 2 O 2 , should be put in glass-stoppered bottles because they 
cause rubber deterioration. 

4. H 2 S Generator. Although directions in this text are given primarily 
on the assumption that an aqueous %% solution of thioacetamide will be^ 
used as an H 2 S sourceJHTaS gas as such can be produced in a generator and 
conducted to the container where precipitation is to be made. Many 
schools have their own design for the H 2 S generator and only a simple one 
is sketched here. If a set up similar to this is employed, the student will 
have to make a number of capillary delivery tubes because they become 
contaminated with sulfides that are difficult to remove. The cartridge in 
Fig. 14-3 contains the commercially available product "Aitch-tu-ess," or a 
self prepared mixture of sulfur, paraffin, and asbestos, either of which 
when heated gives H 2 S gas. If a cylinder of H 2 S or a Kipp generator is 
employed, a scrubbing bottle should be in the line between source and 

In putting glass tubing through holes in stoppers one should lubricate 
it with a little glycerine then wash off the excess with water. As an 
added precaution, tubing may be held through several folds of a towel to 
minimize danger of cutting one's self in case of breakage. 

delivery tube of 

| oose 
alass wool 

AUcli-tu-css " mixture 
and tabe 

FIG. 14-3. One type of student H 2 S generator. 

5. Gutzeit-Fleitmann Apparatus. These tests are further described 
in Chapter 16 under arsenic and antimony. A piece of 8-mm tubing is 
heated and drawn as described in section 1 above, this time to produce a 
tip with about 4-mm O.D. It is cut off to the dimensions given in the 


accompanying figure, fire polished, and fitted snugly to a 3- or 4-in. test 
tube with a single-hole cork or stopper. 


Q-mrn tubing O.3). 
50% AgN0 5 spot 

4-in . tabe 

Cotton t Pb(NOj) 1 to 
remove H?? 


FIG. 14-4. The Gutzeit-Fleitmann apparatus. 

6. Platinum Wire. Seal a 2-in. piece of 24- or 26-gauge platinum wire 
into a piece of glass tubing or rod by heating the glass until soft and push- 
ing the wire into it. Anneal the metal-to-glass seal with a semi-luminous 

FIG. 14-5. (a) Pt wire in a test tube, (b) Nichrome wire with cork handle. 



flame for at least a minute. Assemble the rest of the apparatus as pictured. 
Use is made of this in the first parts of special experiments 4 and 5. 

Nichrome (Ni Cr Fe alloy) wire may be used in most cases as a 
substitute for platinum. When it becomes contaminated it is simply 
discarded. Two-inch pieces are bent into a loop at one end and pushed 
into a small cork at the other. 

7. Wash Bottle. The correct angles for the glass bends are conveni- 
ently made by drawing them on an asbestos board, then bending hot 6-mm 
tubing to fit the pattern, and later cutting to the dimensions. The tip 
is drawn out as described in section 1. Its I.D. should be about 1 mm. 
A 125-, 200-, or 250-ml Erlenmeyer or Florence flask may be used for a 
wash bottle, and heavy string may be wound around the neck for insulation 
in case hot water is the wash liquid. A combination of cork sheet and 
tape works in the same capacity. 

trapping -re qas or 
collecting distillate 

FIG. 14^6. Wash bottle made from 
a 125-ml Erlenmeyer. 

FIG. 14-7. Gas delivery or distillation 

8. Gas Delivery and Small Distillations Apparatus. In the in- 
vestigation of gas-releasing reactions or in carrying out distillations, the 
device in Fig. 14-7 is useful. The reaction mixture is placed in the test 



tube and the delivery tube is placed inside another tube. This second 
tube may contain an absorbing solution, as Ca(OH) 2 , to trap CO 2 or SO 2 , 
or may stand in a beaker of cold water for condensation of vapors, as in 
the distillation of HSCN. 

9. Spoon for Solids. Seal one end of a piece of 4-mm tubing shut 
(1) by heating, and while hot, blow a small bulb on it (2). Direct a small, 
pointed flame to one side of the bulb and when it becomes hot, suck in 
slowly at the open end of the tube to pull in the glass and form a spoon 
(3). Cut the tubing off at a convenient length, say about 10 cm, then seal 
shut that end in the flame (4). 


FIG. 14-8. Steps in blowing a glass spoon. 


Methods of conducting special tests like those involving the blow pipe, 
filter paper spots, borax beads, fusions, etc., are described in the appropriate 
laboratory directions. Some general techniques of using the locker 
equipment are summarized below. 

10. Heating Solutions. Solutions boiled in small test tubes tend to 
superheat and suddenly bump out. This may be avoided by slow, careful 
heating near the top of the liquid if a direct flame is used, though there is 
always still a danger that solutions may "bump" or be inadvertently 
heated to dryness and the residues baked. One can circumvent this by 
the use of air or water baths as illustrated in Fig. 14-9. Such baths are 
purchased or made from beakers of desired sizes with a lead, aluminum 
or copper sheet bent over the top and having in it holes of sizes to fit test 
tubes or micro beakers. Alternatively, a twisted wire can be made to do 
the holding. Rings cut from rubber tubing are put on the tubes as collars 
to fix their positions in the perforated plate, or the test tube holder is left 



on a tube for the same purpose. Rubber bands should be replaced 
regularly as they dry out and become brittle. 


rubber rinq 

iron rinQ 
rinq stand 

FIG. 14-9. Water bath made from a perforated metal sheet bent over a beaker. 
Water is emptied and the vessel is used dry as an air bath. 

Care must be taken with evaporations if the setup is used as an air bath, 
because bumping due to super heating can occur. This is avoided by 
using a low flame and allowing some minutes for the operation. As the 
whole apparatus retains heat after the fire is shut off, evaporation continues 
and the last few drops of moisture in a tube should be driven off in this 
way. Too much heat can change the chemical composition of some 
residues. Evaporations giving noxious fumes are to be done under the 

11. Handling Solids. It is best to handle solids in glass only in order 
to guard against reaction with or contamination from paper or metal 
spatulas. A spatula should not be used to transfer solid chemicals from 
a stock bottle to another container; rather one taps out the material 
on a watch glass and measures with the balance, spatula, or spoon 
what is needed from that, in oider to keep the original chemical 
supply of unquestioned purity. A little practice in quantity estimation 

Small amounts of solids can be pulverized using a short, blunt 8-mm 
stirring rod in one of the depressions of a spot plate. Using a very fine 
carborundum dust on one depression with the rod will give grinding 
surfaces like those ot a mortar and pestle, and solids will not slip out when 



put under pressure. These rough surfaces must be well cleaned before 

The student should note the remarks on the labels of side-shelf solids in 
reference to their reaction with CO 2 and H 2 O from the air and should 
keep these chemicals particularly tightly closed when not being transferred. 

The glass spoon described in section 9 may be calibrated by filling it 
level with a solid like NaCl and weighing the salt. Except with technical 
grade chemicals, the spoon should not be used to remove solids from 
bottles because of the danger of contamination of the main supply, even 
though this smooth spoon can be readily cleaned. 

12. Separations. Mixtures of liquids and solids are separated by 
means oi a centrifuge unless the particulate phase settles rapidly and/or 
only an approximate separation is needed, whereupon simple decantation 
is used. The supernalant liquid is called the centrate or decantate, respec- 
tively, in these operations, and the solid is called the residue. The liquid 
is drawn off as pictured in Fig. 14-10, with the operator being careful not 
to remove it too rapidly or some residue may come with it. If more 
complete separation is demanded, the solid is mixed with wash liquid, 


with tubes 

- centr ate 



FIG. 14-10. : aking 
off the centrate with a 
pipet after centrifnging. 

FIG. 14-11. A centrifuge. 


perhaps heated, then recentrifuged, and the wash liquid removed as 
before. Laboratory directions will be explicit on this. 

Pipets and droppers are never to be used to withdraw liquids from reagent 
shelf bottles. 

A motor-driven centrifuge is recommended, having head openings and 
cylinder depths to fit both 3- and 4-in. test tubes. The solution to be 
centrifuged is balanced on the opposite side of the machine with another 
tube containing the same amount oj solution or water so that the bearings 
will not be worn unduly. Tubes are not to be over two-thirds full. Any 
solution spilled on the centrifuge should be cleaned up immediately and 
if such an accident occurs with corrosive fluids the instructor should be 
notified so he may keep damage to the machine at a minimum. 

A good centrifuge operates at 1500-4000 revolutions per minute (rpm) 
and the usual laboratory type has a radius of 8-10 cm. Maximum speed 
is attained in 15-30 sec and most precipitates arc settled by the centrifugal 
force in 30-60 sec, after which time the switch is turned off and the 
revolving head slowed with gentle brake or hand pressure. The centrifuge 
is a relatively expensive instrument and is not to be mistreated. 

TJie centrifugal force,/, acting on a settling particle of mass, m, is 

f=4n*mRV (14-1) 

where R is the centimeter radius measured from the center of the centrifuge 
to the end of the test tube, and V is the rpm. (Most physics books derive 
this formula.) For purposes of contrast with gravitational induced 
settling it can be shown that the centrifuge's force compared to the 
gravitational force, #, is related by 

/=(1.12x 10-W (14-2) 

For a centrifuge with a 10-cm radius whirling at 2000 rpm its force is 
448 times the gravitational force, or it settles particles in 1/448 the time. 
The term, w, in equation 14-1 is recognized to be governed by the density 
of the solid and the size of the particles. Not considered here is the 
viscosity of the liquid medium and its density as these are not important 
to us. 

13. Washing Equipment. Since only small quantities of materials 
are handled, general cleanliness must be observed with the laboratory 
implements. Dirty equipment should be cleaned between analytical 
operations. For most purposes, a test tube brush and mixture of scouring 
powder and detergent, followed by rinses with tap and distilled water give 
good results. In extreme cases where this fails, a little aqua regia (concen- 
trated HC1 and HNO 3 ) heated under the hood may be tried. A clean 


feather and soap solution can be used to clean pipets and small-diameter 
tubes, though it is frequently more satisfactory to discard stubbornly 
soiled glass tubing and to prepare new bends and jets. 

14. Mixing Solutions. Solutions can be mixed using a stirring rod in 
a test tube of solution or by holding the test tube lip between the index 
finger and thumb of one hand and tapping the lower part of the tube with 
the index finger of the other hand. 

Caution should be used in mixing solutions which are concentrated, 
contain strong oxidants, or are hot. One should add acids and bases to 
water, rather than the reverse, to avoid splattering. Add small amounts 
with stirring to see what will happen before mixing larger quantities. 
Point the container away from everyone. 

Preliminary Experiments 

Chapters 15-22, inclusive, deal with the analysis of 32 cations and 
18 anions. Of the cations, 24 have been analyzed in beginning qualitative 
courses for years and schemes of separation and identification are well 
known and are given here as directions for separating and identifying 
mixtures of their compounds. Before the unknown sample is run with 
these procedures, however, it is instructive to perform several preliminary 
tests on the ions to be analyzed in order to characterize their chemical 
behavior as an aid in recognizing them in the samples. One should 
also get from the preliminaries the reaction equations, the evidences to 
look for which constitute positive tests, correlations among similarly 
acting ions, and something of the sensitivities and interferences of the 
supplementary tests. Some of the tests are general for a group of 
ions while some are specific for a single ion. When the latter follow 
positive reaction with the general reagent, they are called confirmatory 

The test solutions on which preliminary investigations are made contain 
10 mg of the ion under analysis per milliliter of solution, so one drop 
contains about 0.5 mg of that ion. By running tests on various dilutions 
one is able to compare quantities of precipitates or depths of colors 
obtained with the sample and so estimate quantities in the sample as small, 
medium, or large. He is also able to determine in this way the sensitivity 
of a given method or reagent and to compare it with other tests. Directions 
for a number of these opportunities are given later. Prior to running an 
unknown, one may, if desired, go through a known mixture (made from 
the test solutions) according to the procedure. If one performs the 
preliminary studies carefully, takes notes on them and understands their 
relation to the equations and procedures which follow, he has little trouble 
with the unknown sample. It is common practice to have an unknown 


for each group of chemically similar ions and then a general unknown 
that may contain any combination of them. 

The discussion of eight "less familiar" metals in Chapter 20 does not 
include a procedure for separations, although preliminary tests are given. 
These tests are sufficient to identify the elements in their usual alloys 
and matrixes. 

Chapter 22 on anion analysis does not contain any general separation 
procedures beyond the use of three screening reagents, since distinctive 
evidences are available as outlined in the preliminary tests to identify any 
of the anions in the presence of any others. One then reuses these tests 
to analyze the unknowns. 

In making any analysis, it is safest to compare results with a check of a 
known sample from the shelf test solutions, and for specific tests it is wise 
to run a blank to see if unknown factors such as contamination or a 
deteriorated reagent have given false indications. A blank is a test in 
which everything but the ion in question is included to give a survey of the 
other factors present. 

Chapter 23 is a compilation of special experiments to illustrate principles, 
special methods, and applications of qualitative testing. 

The Laboratory Notebook 

Organization of a laboratory notebook is dictated by the instructor's 
preference. The following is one idea. 

A separate notebook is used exclusively for laboratory directions, 
observations, and results, and is brought to the laboratory each time. An 
inexpensive spiral bound book containing about 50 sheets of 8| X 11 in. 
paper is satisfactory. For each group of ions studied, the student writes 
in his own words the procedure he will use on the sample. This is done in 
abbreviated fashion from the text directions and his experience with 
preliminary tests in such clarity that he can analyze samples from his 
notes alone. This is not an exercise in penmanship but an evidence that 
the student really understands the unit of work involved, since he must 
select for outline all the important ideas. Following the procedure are 
written observations on the preliminary laboratory work, lecture notes 
that pertain to the same thing, and answers to laboratory questions that 
might be chosen from the text or another source. This work being 
completed, the student is well prepared to tackle the sample. If each 
person organizes his book in the same way, including numbering of 
numbered sections and experiments from the text, the notes are easily 
checked and graded. 

Reporting the unknown sample results is again a matter of the instruc- 
tor's preference. If separate report forms are not used then a space, say 



3x5 in., is reserved in the notebook. It might look like this in 
simple form: 

Name ._ 

Sample No 

Ions found and 
quantities estimated 

Lab sect. 


Checked by 

FIG. 14-12. One form of unknown report. 

More detailed report forms may call for the results from certain procedural 
steps, reagents used, colors of important precipitates and solutions, 
reaction equations, and conclusions. 

Survey of Cation Separations 

A considerable portion of the laboratory time is usually devoted to 
analysis of 24 cations (including the two oxidation states of mercury and 
the ammonium ion), and a brief look at this segment of work is helpful in 
orientating the student. 

If one has a mixture of most any combination of these 24 ions, the 
flow sheet given on p. 229 presents a scheme for separating them into 
five groups for further investigation, by use of reagents that precipitate 
only certain ions under given concentration and temperature conditions. 
Thus, if dilute HC1 is added to the general solution mixture, Ag*, Hg+ 2 , 
and Pb^ 2 precipitate as chlorides and are segregated by centrifugation. 
These metals are called group 1. The /?H of the supernatant liquid is 
adjusted to about 0.6 and H 2 S or thioacetamide, CH 3 CSNH 2 ( a source of 
H 2 S), is put in and group 2 metals precipitate as sulfides. These include 
Hg+ 2 , Pb+a, Bi' Cu> 2 , Cd< 2 > As nj , Sb" T , and Sn 11 . (The mercuric 
mercury is not obtained in group 1 and lead chloride is not completely 
precipitated, so these metals may be detected in both groups, although only 
the mercury is to be reported as a different ion because the two oxidation 
states have different properties.) The centrate from the precipitation 


TJ D r . 


pa u u 

Q> u, W) c/> </) JQ JD C C 


I . 


2 < 

1 L 

S r^ "*" r ^ 
/3 QJ 3 ' "" ^ 


CJ So 





1 s 


u- b 

o " 

| 1 1*^ t ^ 


X sj_ 








S o., 













u 5 




u s, 

CA o> 3 

1 o 

S c oc 





again contains ions of subsequent groups and is saved for their separation. 
The residue of group 2 sulfides is divided chemically into sub-group 2A, 
sometimes called the "copper group," and subgroup 2B, also called the 
"tin group," by addition of 6 M KOH, which dissolves the amphoteric B 
group sulfides of As, Sb, and Sn and leaves the other sulfides unaffected. 
The centrate from the original group 2 precipitation is buffered with an 
ammonia system, and (NH 4 ) 2 S is added. In this basic solution the 
seven ions of group 3 precipitate, Fe 43 , J+ 3 , and Cr' 3 as hydroxides, 
Mn f2 , Zn+ 2 , Co f2 , and Mi*" 2 as sulfides. **" 

The group 3 residue is treated with NaOH and H 2 O 2 to divide it into 
two groups: subgroup 3B, the "aluminum group," which includes Al, 
Zn, and Cr in solution and leaves behind a residue of the others as sub- 
group 3A, the "iron group." The centrate from the original precipitation 
of group 3 is treated with (NH 4 ) 2 CO 3 in an alkaline buffer solution which 
throws out the group 4 ions, Ba+ 2 , Ca +2 , and Sr f2 , as carbonates. The 
centrate from this contains Mg^ 2 , Na f , IC f , and NHJ, ions which are 
called group 5 and need no group precipitating agent. (Since NH was 
added as ammonia or its salts in the procedure, it will always be present in 
group 5 of the general unknown, so the original sample is tested for this 
ion.) The five groups are subjected to further separations and their 
components verified. Detailed directions and equations are given in the 
ensuing chapters for all necessary steps, and summaries in the form of 
flow sheets are also presented to give the over-all pictures in minimum 



The three heavy metals of group 1 are taken together as an analytical 
unit because of the low solubility of their chlorides, although they are in 
three different groups of the periodic table as indicated: 



























FIG. 15-1. The periodic table in the vicinity of the group 1 metals. 


This element has the highest heat, electric conductance, and light 
reflectivity of any metal and is second only to gold in ductility. It is 



found free in nature as well as in sulfide (silver glance) and chloride (horn 
silver) minerals. The metal is dissolved from ores by sodium cyanide 
solutions in which soluble silver cyanides form. They are later decom- 
posed with zinc: 

AgCl + 2CN- = Ag(CN)- + Cl- 

The potential is 

Ag + 2CN- = Ag(CN)~ + e~ E = 0.29 

The metal is used in mirrors, dinnerware, plating, jewelry, and coinage. 
Silver halides are photosensitive and used in photographic emulsions. 
Silver is fairly inert but will dissolve in nitric acid, alkali cyanides, and 
molten strong alkalis in the presence of air. Its sp. gr. is 10.5, its b.p. is 
1940 C and m.p. is 960.5 C 

The oxidation states of the metal are (I), (II), and (III), as are those of 
the other coinage metals, Cu and Au, but only (1) is important. The 
higher states can be produced by anodic, peroxydisulfate, or ozone oxida- 
tions of Ag f , three methods which are common in research on higher 
oxidation states of many elements. In nitric acid for example 

Ag + + 3 = AgO+ + 2 (slow) 
Ag*- + AgO+ + 2H+ = 2Ag Tr + H 2 (f as t) 

These unfamiliar oxidation states have large negative potentials as 

expected : 

Ag+ + H 2 = AgO+ + 2H+ + 2e~ E = - 2.0 

Ag' 2 + H 2 = AgO+ + 2H^ + e~ E = - 2.1 

which indeed are so large that they decompose water with the evolution of 
oxygen. Ag TI coordinates four groups. 
The Ag 4 ion is colorless and easily reduced. The half reaction is 

Ag = Ag+ + er E = - 0.80 

It forms many compounds of low solubility and many stable complex 
ions having 2-coordination. Table 15-1 lists some of the very slightly 
soluble silver compounds in order of decreasing equilibrium [Ag+]. 
Among the halides, AgF is peculiar in being quite soluble; its saturated 
solution is about 14 M at room temperature. Silver nitrate is the most 
common silver compound and its solubility is also high, 3000 g dissolving 
in a liter of H 2 O at 30 C. It is made by the action of nitric acid on the 
metal. Silver nitrate does not dissociate strongly in water and is appreci- 
ably soluble in many organic liquids. With strong bases, Ag 4 " gives 
Ag(OH), which spontaneously changes to black, insoluble Ag 2 O, an oxide 



with alkaline reactions strong enough to indicate that silver salts are not 
going to be hydrolyzed greatly. Among other silver compounds worthy 
of mention, the nitrite, sulfate, and acetate are of moderately low solubility, 
and some of the other silver salts are of interest because their colors give 
the analyst a clue to their identity: Ag 3 PO 4 is yellow, Ag 2 CrO 4 dark red, 
and Ag 4 [Fe(CN) 6 ] -H 2 O yellow. 

Soluble complexes other than the cyanide already mentioned are those 
with halides, ammonia, and thiosulfate, as AgClj, Ag(NH 3 )+, and 
Ag(S 2 O 3 ) 2 3 . The first accounts for the fact that AgCl is considerably 
more soluble in moderately strong HC1 solutions than in H 2 O, the common 
ion effect notwithstanding; the second is of analytical importance because 
of its formation when AgCl from the group 1 precipitation is solubilized 
in NH 4 OH; and the third has practical value because it is the soluble 
product from treatment of exposed film (containing urireacted silver salts) 
with sodium thiosulfate (hypo). 

In the qualitative analytical procedure, silver is precipitated as the 
chloride, dissolved as the ammine complex, and reprecipitated by nitric 
acid, which breaks up the complex and permits recombination of silver 
and chloride ions: 

AgCl + 2NH 3 = Ag(NH 3 )< + Ch 
Ag(NH 3 ).J- + Cl- + 2H f + 2N0 3 - = AgCl + 2NH| + 2NO 3 

The following table summarizes silver reactions with regard to slightly 
soluble compounds and slightly dissociated Werner ions. The order w all 


Reactions (in order of decreasing [Ag+]) 


Formula Weight 

AgAc = Ag+ -f Ac~ 

2.3 x lO- 3 


AgNO 2 = Ag f + NO^ 

1.2 x 10- 4 


Ag 2 C0 3 = 2Ag+ + CO," 2 

8.2 x 10~ 12 


AgiO 3 = Ag+ -f lOg- 

3 x 10 8 


Ag 2 CrO 4 = 2Ag f + CrO 4 2 

1.9 x 10-" 


AgCl = Ag+ + Cl 

2.8 x 10- 10 


AgSCN = Ag f + SCN- 

1 x 10- 12 


AgBr = Ag+ -f Br" 

5 x 10 13 


AgCN = Ag< + CN~ 

1.6 x 10~ 14 


Ag(NH 3 ) 2 f = Ag+ + 2NH 9 

5.9 x 10- 8 


Agl = Ag f + I- 

8.5 x 10- 17 


Ag(S 2 3 ) 2 3 = Ag+ -f 2S 2 O3-2 

6.0 x 10- 14 


Ag(CN)a- = Ag+ + 2CN- 

1.8 x 10' 19 


Ag 2 S = 2Ag+ -f S' 2 

1 x 10- 50 



following compilations of this type is that of decreasing equilibrium molar 
concentration of the ion under discussion. Thus Ag 2 S, having the least 
[Ag+] in equilibrium with it, will form by the addition of S~ 2 to any other 
silver compound or ion, etc. For the purpose of these tables, complex ions 
and complexing agents will be taken as 1 M and the metal ion calculated 
accordingly. Thus, if a solution of Ag(NH 3 ) is 1 A/, and the [NH 3 ] in the 
solution is also 1 M [Ag+] is 5.9 X 10~ 8 M. 

Silver is analyzed quantitatively by both gravimetric and volumetric 
means. It is either precipitated as the chloride, dried at HOC, and 
weighed as AgCl, or the Ag + is titrated with standard thiocyanate (Volhard 
method). Ferric ion is used as an indicator in the latter red Fe(SCN) +2 
forms when the silver has just been quantitatively precipitated as AgSCN. 


Lead is a soft, malleable, not very ductile metal. The ore, galena, 
PbS, is roasted to provide PbO which is carbon reduced. Lead is used in 
radiation shielding and on electric cables and in important soft, low- 
melting mixtures with Sn, Sb, Bi, and Cu to form such alloys as pewter, 
battery plate, shot, solder, white metal, rose metal, and type metal. 
Important industrial lead compounds are the sulfate, chromate, basic 
carbonate, and oxides which are used in paints, and the silicate which is 
used occasionally in glass and ceramic manufacture. The metal's jp. gr. 
is_ 1 1.337. m.p. is 327.4 C. andjrjg.jfi 1 3 C. 

Two oxidation states of lead are known, {II) and (IV). Oxides of both 
are amphoteric, leading to four series of lead compounds: Pb+ 2 ,plumbous; 
Pb lv , plumbic; plumbites possibly as PbfOH)^ 2 or as PbO~ 2 and HPbO^ 
in less basic media; and plumbates as HPb(OH)g . Pb IV compounds can 
be prepared by anodic oxidation. The most familiar of these is PbO 2 , 
which is pressed on to storage battery plates. The battery discharge 
reaction is 

Pb + Pb0 2 + 4H^ + 2S(V = 2PbS0 4 + 2H 2 O 

and the half cells from which one can calculate an emf for the whole cell are 

Pb + SO^ 2 = PbSO 4 + 2*r E = 0.36 

PbS0 4 + 2H 2 = Pb0 2 + SOj 2 + 4H^ + 2e~ E = - 1.68 

Pb lv compounds are easily hydrolyzed and reduced in water solution, and 
as far as analysis here will be concerned, only Pb 4 * 2 compounds will be 

Pb+ 2 is colorless and stable in water solutions below pH 7. Between 
pR 1 and 13, Pb(OH) 2 (hydrated) is formed, and above 13, plumbites 
result. The acetate, chlorate, and nitrate are soluble though they ionize 



but feebly and hydrolyze (unless some acid is present) to precipitate basic 
salts of variable composition nominally written as Pb(OH)Cl, etc. Most 
lead salts have low solubility. Lead does not form complexes as readily 
as does silver, and other than the hydroxides listed above, the chloro 
complexes like PbClj 2 will be the only ones mentioned here. 


Reactions (in order of decreasing [Pb+ 2 ]) 


Formula Weight 

PbCl+ = Pb+ 2 + Cl- 

7.75 x JO" 1 


PbCl 2 = Pb+ 2 + 2C1- 

1.6 X 10- 5 


PbBr 2 = Pb+ 2 + 2Br~ 

4.6 x 10 6 


PbF 2 = Pb f2 + 2F- 

4 x 10~ 8 


PbI 2 = Pb+ 2 4-21- 

8.3 x 10" 9 


PbSO 4 = Pb +2 4- SO 4 2 

1.3 x 10- 8 


Pb(OH) 2 = Pb^ 2 + 20H 

4 x 10~ 15 


PbC 2 4 = Pb+ 2 -h C 2 4 2 

8.3 x 10~ 12 


PbCO 3 = Pb+ 2 + CO^ 2 

1.5 x 10^ 13 


PbCrO 4 = Pb+ 2 + CrO 4 2 

2.0 x 10- 16 


Pb 3 (PO 4 ) 2 = 3Pb+ 2 + 2POi 3 

1 x 10- 54 


PbS = Pb+ 2 4- S~ 2 

4 x 10- 26 


Lead is gravimetrically determined in quantitative analysis by oxidation 
to PbO 2 at a Pt electrode from HNO 3 solution or by precipitation and 
weighing as PbSO 4 . Dithizone is a very sensitive color reagent for lead 
and used to measure it quantitatively in concentrations even below 
1 ppm. (See special experiment 10.) 


Mercury is the only metallic element which is a liquid at 25 C and one 
atmosphere (melting points of other low melting metals are Cs = 28.5 C, 
Ga = 29.75 C). The common ore is cinnabar, HgS, which is roasted to 
give the oxide, which in turn decomposes with heat yielding oxygen and 
mercury. The metal is commercially distilled for purification in the 
presence of air which oxidizes usual impurities of more active metals like 
Cd to drosses that are left behind when Hg is vaporized. Washing in 
dilute HNO 3 followed by a vacuum redistillation yields almost 100% 
pure Hg. Itsjp. gr. is 13.546, m.p. is 38.87 C, and b.p. 356.95 C 

The metal has a near linear expansion with temperature which malces it 
useful in thermometers. It is also used in precision casting to make 
unusually smooth ceramic molds for the production of close tolerance 


radar and jet engine components. Hg conducts electricity relatively well 
in the liquid state and is utilized in switches; it also conducts in the vapor 
(monatomic) state, the mercury vapor lamp giving a brilliant bluish light. 
Most metals, Pt, Co, Ni, and Fe excepted, dissolve in mercury. The 
liquid mixtures are called amalgams (some are true solutions, others 
suspensions) and use is made of them in gold and silver recovery and in 
dental fillings. Mercury compounds do not have wide utilization; all 
the soluble ones are poisonous. Mercurous chloride, calomel, is used as a 
medicine for liver stimulation. Mercuric Julminate, Hg(CNO) 2 , is used 
in caps to set off dynamite. 

The two oxidation states are (I) mercurous and (II) mercuric. There 
is probably no Hg+; instead, mercurous mercury exists in the solid com- 
pounds and in solution as HgJ 2 . See problem 13, p. 246. This doubled 
ion is possible because the electron structure of the Hg atom is complete 
through the 5d shell and has two more electrons in the 6s. Loss of one 
of the 6s electrons and coupling of two such ions by sharing their remaining 
6s electron gives mercurous ion the structure +Hg:Hg 4 \ There are 
various methods of proving this from (a) equilibrium constant evaluation 
experiments as explained in Chapter 9 for the I~ ion; (b) conductivity 
comparison with other heavy metal salts of both possible types MX 2 and 
MX; and (c-) X-ray studies on the crystal structure of mercurous chloride. 
The last method shows, for example, that the arrangement is 
( Cl Hg Hg Cl ),, whereas in silver chloride the structure is 
( Cl Ag C*! Ag ) je . Most of the reactions of mercurous salts are 
those of reduction to the metal or oxidation to Hg f2 , The potentials are 

2Hg = Hg*- 2 + 2e~ = - 0.79 

Hg 2 = 2Hg 2 + 2r- E = - 0.92 

Hg = Hg+ 2 + 2e~ = -0.85 

These show that reductions from mercuric to mercurous ion and to the 
free metal are easy, a fact utilized as a test for the element. 

The mercurous salts are freely ionized and HgJ 2 forms few complexes. 
Most mercurous salts have low solubility. Hg 2 Q 2 is the most common 
mercurous compound and is precipitated in cation group 1 ; it is more 
soluble than the bromide* iodide, or sulfide. The common soluble mer- 
curous salts are the acetate, nitrate, and sulfate, but Hg 2 hydrolyzes 
slowly and basic salts are prone to precipitate from these solutions. 
Adding strong base to Hg 2 results in precipitation of Hg 2 O, which on 
prolonged standing disproportionates to HgO + Hg. 

Mercuric compounds resemble those of lead and cadmium in that most 
are only a few per cent dissociated at moderate dilutions, and the amount 



varies with the anion present. The Hg +2 ion hydrolyzes in water since 
Hg(OH) 2 is a weak base giving basic (hydroxy) salts, and ammonolyzes 
with ammonia in similar fashion to give basic (amido) salts. Hg+ 2 forms a 
soluble tetraammine complex but it is stable only in concentrated NH 4 OH. 
A salt both ammonolyzed and hydrolyzed is HO Hg NH Hg I, 
the product from Nessler's test (see Chapter 16, Hg+ 2 tests). Mercuric 
sulfide is the most insoluble mercury compound, but it will dissolve 
appreciably in alkaline sulfides or polysulfides to give HgS^ 2 , and dissolves 
completely in aqua regia, giving HgCl 4 2 . Hg 11 is found in a number of 
soluble complexes as listed below; the coordination number is 4. Normal 
salts which are soluble include HgCl 2 , therefore Hg+ 2 doesjiot precipitate 
as ajjrqup 1 chloride. 

In the qualitative scheme of testing, both mercury valences, as well as 
the free metal, are involved, since when mercurous chloride is precipitated 
and reacted with ammonia, mercuric amido chloride and mercury are 
products : 

HfeCl^ + 2NH, = Hg(NH 2 )CI ( , + Hg, + NH{ + Cl~ 

These valence >tatc > also appear in the reaction between (excess) stannous 
ion and mercuriv \m later in the analytical procedure: 


+ 2 

+ Sn 11 
Hg 2 Cl 2 

2C1- = 
Sn 11 = 2Hg 

+ Sn lv 
Sn lv + 2CI 


Reactions (in order of decreasing [Hg.; 2 J) 


Formula Weigh! 

Hg 2 Cr0 4 = Hg 2 ia + rr0 4 2 

2 x 10" 


HggCJIA - Hg 2 < 2 + C.HA"* 

1 x 10 1() 


Hg 2 C 2 O t = Hg 2 i2 + C 2 4 a 

J x lO' 13 


Hg 2 Cl 2 =Hg 2 f2 + 2Cl- 

1.1 x 10 18 

472 .1 4 

Hg 2 (SCN) 2 = Hga* a + 2SCN- 

3 x 10- 20 


Hg 2 Br 2 = Hg,,' 2 + 2Br - 

1.3 x 10- 22 


Hg 2 CO, = Hg 2 2 + C0 3 2 

9.0 x 10~ 17 


Hg 2 I 2 = Hg L + 2 + 21 

4.5 x 10 29 


Hg 2 (C'N)o = HgJ 2 -f 2CN- 

5 x 10- 40 


Hg 2 S = Hg+ + S- 2 

1 x 10 45 


Note again the conventions given above Table 15-1 regarding how 
metallic concentrations from complex ions will be calculated. 



Reactions (in order of decreasing [Hg+ 2 ]) 



Hg(I0 3 ) 2 = H 

g t2 + 2I0 3 - 

3 x 10- 13 


HgCV = Hg+ 2 + 4C1- , 

8.3 x 10~ 16 


Hg(NH 3 ) 4 + 2 = 

Hg t2 -f 4NH.j 

5.2 x 10- 20 


Hg(SCN>4 2 = 

Hg f 2 + 4SCN 

5 x 10- 20 


HgBr 4 2 = Hg 

^ 2 + 4Br~ 

2.3 x 10~ 22 


HgS = Hg^ 2 - 

h S^ 2 

1 x 10- 50 


HgI 4 2 = Hg+ 2 

+ 41 

5.3 x 10- 31 


Hg(CN) 4 2 = } 

lgf2 + 4CN ^ 

4 x JO" 42 


In volumetric quantitative analysis, one titrates with SCN~ using Fe f3 
as an indicator: 2SCN~ + Hg+ 2 = Hg(SCN) 2 . In the gravimetric 
procedure, one uses a weighed gold foil with which mercury vapor 
amalgamates in a specially designed heating apparatus. 

Group 1 Analysis General Discussion 

To a neutral or slightly acid solution, 6 M HC1 is added dropwise until 
precipitation is complete, then a drop in excess. The three group 1 ions 
precipitate as white chlorides, AgCl, PbCI 2 , and Hg 2 Cl 2 . The centrate 
from their separation is drawn off and saved for groups 2-5 analysis if 
those ions might be present. Water is added to the residue, the mixture 
is heated, and recentrifuged. PbCl 2 is more than three times as soluble 
at 100 C as at 20 C and it is solubilized and separated by this means, leaving 
AgCl and Hg 2 Cl 2 as a residue. The centrate may be divided into two 
portions: H 2 SO 4 added to one, K 2 CrO 4 to the other. If lead is present, 
white PbSO 4 and yellow PbCrO 4 will furnish the evidence. The residue 
of silver and mercurous chlorides is mixed with NH 4 OH, resulting in 
formation of soluble, colorless Ag(NH 3 )+ and a grey-black insoluble 
mixture of mercury disproportionation products: white HgNH 2 Cl and 
black Hg, The mixture is centrifuged, and the silver-containing centrate 
isolated and acidified with HNO 3 . This decomposes the complex and 
Ag f and Ci again combine as shown by a white precipitate which darkens 
to metallic silver in the sunlight. The black mercury mixture is proof 
enough that that element was a constituent of the sample but it may be 
confirmed by solution in aqua regia and reduction to white Hg 2 Cl 2 plus 
black Hg by stannous tin.* 

* A summary of this appears on page 239 in the form of a "flow sheer and on the 
same page in an alternate "block outline." Subsequent groups will be treated with 
description and flow sheet only. 

Group I test 
solution or 

a general 




to be 
treated for 




Group I 


centrifuge other groups PbC\ 2 J) centrifuge 

.. . HO, heat, 
+ Hg 2 CI 2 J '_ > Hg a CI 2 

Residue Centrate 

+ PbCl 2 

Sunlight HN0 3 Centrate 

AgCI * Ag(NH 3 ) 2 + 


Residue yH 4 OH, 
\HgNH 2 CV centrifuKe 

PbS0 4 PbCrO ir 

iHN0 3 , 
HC1 3 

SnClr 8 

Hg 2 CI 2 + Hg 

* Subscript letters refer to the color of the substance. The abbreviations are listed in 
Appendix A27. Colors will be given only once per page or per group of related equations. 


Add 6 M HC1 dropwise to the sample with stirring, until precipitation 
is complete, then add a drop more. Centrifuge, wash the residue with a 
little 1 M HC1 and combine washings with original centrate, and save for 
groups 2-5. Proceed below with the residue of group 1 chlorides. 

Residue: AgCI, PbCl 2 , Hg 2 Cl 2 . Add H 2 O and boil. Centrifuge. 
Residue: AgCI, Hg 2 Cl 2 . Add 6M Centrate: PbCl 2 . Divide into two 

NH 4 OH and centrifuge 

portions, A and B 



A. Add a few 

B. Add a little 6 M 

HgNH 2 CI, Hg. 

Ag(NH 3 ) 2 Cl.Add 

drops of NH 4 Ac 

H 2 SO 4 . A white 

This is sufficient to 

6 M HNO 3 until 

and K 2 CrO 4 ; a 

precipitate of 

confirm mercury. 

acidic. A white 

yellow precipitate 

PbSO 4 confirms 

precipitate of 

of PbCrO 4 con- 


AgCI, turning 

firms lead. 

dark in the sun- 

light, confirms 


t This method of giving a condensed view of the analysis is presented as an alternate 
to the flow-sheet method. The outline form may be used in the student laboratory 
notebook by adding the quantities of reagents described in the procedures for the 



Test solutions contain 10 mg of the metal per milliliter unless stated 
otherwise. All laboratory solutions are listed in the appendix. Most of 
these tests are best conducted in 3- or 4-in. test tubes. Results and answers 
to questions are to be recorded in the laboratory notebook, using the 
numbering system below. Reaction equations for the systematic analysis 
follow this section. 

Test 15-1. Precipitation of AgCl; Effect of Concentration, (a) 

Decide how to make 2 ml of each of the following Ag+ solutions and prepare 
them in separate test tubes: 10 mg/ml, 5 mg/ml, 1 mg/ml, and 0.2 mg/ml. To 
each add 1 drop of 6 M HC1, mix briefly, and note the result both as to quantity 
of ppt., curdy character of the residue, and rate of settling. Centrifuge by 
balancing the tubes opposite each other in the centrifuge and again note the vol. 
of the solid phase. 

By preparing known dilns. in this way, one is able to estimate by residue size, or 
sometimes color intensity of the liquid, the concn. represented by the ion under test 
in an unknown soln. If on the other hand one wishes to determine the sensitivity 
of a method, he makes dilns. to determine the one where the positive test result is 
just capable of detection. Such techniques will be valuable all semester. 

Save the first standard for Test 1 5-2. 

Test 15-2. Solution and Re pre dpi tat ion of AgCl; Confirmatory 
Reactions. Throughout the lab work with unknown samples, the student will 
find the need to check first indications or inconclusive experiments with supple- 
mentary investigations known as confirmatory reactions or tests. The following 
examples of this idea illustrate some methods of proving that the curdy ppt 
from test 15-1 is AgCl. 

(a) To the test 15-1 residue, add 10 drops of dil. NH 4 OH and mix well. 
Explain, using equation 4 in the list of equations at the end of the chapter. 

(b) To the soln. from (a), add a drop of phenolphthalein, then add dil. HNO a 
until the indicator changes, and then 2 drops in excess. See equation 5. 

(c) Allow the residue from (b) to stand in the sunlight. The photochemical 
reduction of Ag+ to black Ag is another confirmatory test. 

Test 15-3. PbCL 2 ; Character of Precipitates. The student should 
observe not only the color and quant, of ppts. but also their physical appearance. 
The slower a ppt. forms, the larger are its particles generally, and some well 
defined crystalline ppts. may be identified by appearance alone under a lens. 
Whereas AgCl and Hg 2 Cl 2 are curdy, PbCl 2 forms nice crystals. 

Precipitate some PbCl 2 , centrifuge, and decant the liq. Add 1-2 ml of H 2 O, 
heat to boiling, and allow to cool. If pptn. does not reoccur, boil off some of 
the H 2 O and let the tube cool again. Note the difference between the appearance 
of AgCl from test 15-1 and this prep., both of which are usually incompletely 
described simply as white ppts. If a lens or microscope is available, examine 
and describe the crystals more completely. 


Test 15-4. Some Hg 2 Reactions; Blank Tests. A blank test is one 
which is run as a check on solvents, reagents and conditions and includes everything 
but the substance being tested. This allows one to determine that no unknown 
factors such as contaminated reagents are giving false analytical results. The 
student will use this technique from time to time. 

/ Add a few drops of dil. HO to a few drops of Hg.+ 2 test soln. Now add 
a few drops of dil. NH 4 OH and several in excess to make the mixt. basic. 
* centrifuge,* discard the centra te, and warm and stir the residue with a mixt. 
of 12 drops of coned. HC1 and 4 drops of coned. HNO y . When the ppt. has 
dissolved, transfer to a beaker and evcip. in the hood, just to dryness, using a 
moving flame to avoid splattering. Cool. Add 10 drops of H 2 O, 6 drops of 
SnCl 4 2 (SnCl 2 in HC1), and note the result. See the mercury equations toward 
the end of the chapter. 

If one now wants to check to find if the reagents alone were capable of giving 
any of the changes noted, and hence would give a false test for Hgj 2 with no 
mercury present, he will omit the Hg* 2 , mix the reagents in the same order as 
before, and discover that without mercurous ion, no ppts. are obtained, thus 
clearing the reagents and procedure of suspicion. 

Test 15-5. Group 1 Chromates; Checking Against Knowns. A 

good check on a mixture or single substance is to obtain the same from the shelf of 
ion test solns. and run whatever investigation is necessary to show that the known 
and unknown are the same or different. 

Put 5 drops of Ag~* soln. in one tube, 5 of Pb } 2 in another, and 5 of HgJ 2 in 
another. Label the tubes. Ask another student to put 5 drops of one of the 
group 1 ions in a fourth tube as an "unknown". Add 3 drops of K 2 CrO 4 to 
the unknown and note the color. Now add 3 drops of K 2 CrO 4 to each known. 
This simple comparison test method will be used many times. 

Test 15-6. Other Group 1 Precipitates; Devising Original Tests. 

It is possible that the enterprising student who is familiar with the chemistry in the 
flow sheets, and looks over such tables as the one in the back cover , can devise 
shot t cuts in the procedure and originate tests not described here. Such methods 
are not to be relied upon exclusively * of course , since there is no substitute for a 
careful following of the formal procedure. The student should consult the instructor 
before using powerful oxidize rs like AT/O 3 , cone. HNO.^ Na. A O& etc. in original 

As described in test 15-5, the following may also be of aid in differentiating 
group 1 ions: NaHPO 4 , KI, NaF, and NaOH. As time permits, try 1 drop 
of reagent with 5 drops of test ion solution, then add 6 more drops of reagent 
with good mixing, and record the results with equations. 

Analysis of a Known Mixture 

Upon the instructor's directions students may be given, or may them- 
selves prepare, a known mixture of group I ions for analysis prior to 

* * This symbol refers to part (C7), p. 244, of the procedure with the unknown sample 
as a starting point for the Hg confirmatory test. 


running the unknown sample; or they may be directed to proceed directly 
to the unknown after completion of any or all of the preliminary experi- 
ments. At this point, the notebook write up should be complete and the 
student should have read through the following procedure and have the 
outline of it in mind. Any questions the student has on the group thus 
far should be answered, since the same problems may arise with che sample. 


A. Chloride Precipitations 

(1) Ag+ + Cl- = AgC\ w 

(2) Hg+ 2 + 2C1- = Hg 2 Cl 2(v 

(3) Pb+ 2 + 2C1- = PbCl 2 ^ 

B. Silver 

(4) AgCl + 2NH 4 OH = Ag(NH 3 ) + Cl" + 2H 2 O 

(5) Ag(NH 3 )+ + Cl- + 2H*- = 2NH+ + AgCl 

(6) 2Ag> + CrOj 2 = Ag 2 Cr0 4/? 

C. Mercurous Mercury 

(7) Hg 2 Cl 2 + 2NH 4 OH = Hg(NH 2 )Cl H , + Hg, + NH+ + Cl- + 2H 2 O 

(8) Hg(NH 2 )Cl + 4H+ + 6C1- + 2NOj 

= 2HgCl 4 2 + JV 2 + 2NO + 4H 2 O 
3Hg + 8H' + 12C1~ + 2NC>3 

= 3HgCl 4 2 + 2NO + 4H 2 O 

4H+ + 3C1- + NOf = NOCl + C/ 2 + H 2 O (aqua regia) 
2NOCI = 2NO + C/ 2 

(9) 2Hg+ 2 + SnClj 2 + 4C1- = Hg 2 Cl 2)v + 

Hg 2 CI 2 + SnClj 2 = 2Hg + 

(10) Hg + Cr<V = Hg 2 Cr0 4or 

D. Lead 
(11) PbCI 2 + H 2 = Pb* 2 + 2C1- + H 2 O (some ionization) 

(12) Pb+ 2 + SCV = PbSO iw 
PbSO 4 + 2Ac~ = PbAc 2 + SOj 2 

(13) Pb+ 2 + CrO^ 2 = PbCr0 4 , 



(A) Put 1 ml of sample at about pH 1 in a 4-in. tube and add 2 drops 
of 6 M HC1. Shake or stir and allow it to stand a half minute to complete 
the precipitation and make sure that PbCl 2 supersaturation has not occurred. 
No precipitate at this point indicates that Ag+ and HgJ 2 were not in 
the sample and that only a small amount of Pb+ 2 is possibly present. (If 
this is the observation on a sample that may contain only group 1 ions, a 
separate neutral milliliter portion is tested for lead by adding a drop of 
NH 4 Ac followed by 2 drops of K 2 CrO 4 . A yellow turbidity or fine, 
slowly settling precipitate is PbCrO 4 . Lead is confirmed in the PbCrO 4 
by centrifuging, isolating the residue, and warming it with a few drops of 
dilute HC1 in which lead chromate is soluble. If the original sample was a 
general one that might contain cations of other groups, the milliliter 
acidified with HC1 and yielding no group 1 precipitate is carried to the 
group 2 procedure in the next chapter. Any lead not isolated as the 
chloride will be precipitated as the sulfide there.) 

(B) If a white precipitate was the result of HC1 addition in (A), the 
tube is counterbalanced with another in the centrifuge and centrifuged 
for 20-60 sec. The centrate is withdrawn and saved in a corked tube for 
further group analysis if such is indicated, or is neutralized with NH 4 OH, 
buffered with NH 4 Ac, and tested for traces of lead with chromate, as 
described in (A) if the sample is known to contain only group 1 metals. 

(C) The white precipitate from (B) may be any combination of AgCl, 
PbCl 2 , and/or Hg 2 Cl 2 . Wash the solid with 4 drops of 1 M HC1, centrifuge, 
and discard the centrate. Add 1.5 ml of H 2 O to the residue, place the 
tube in a water bath, and keep it hot for 3-4 minutes, stirring once in a 
while with a rod resting in the tube. If all the residue dissolves, only 
Pb+ 2 was in the sample. If some residue remains, centrifuge while hot 
and put the centrate in a separate tube. Wash the residue with 1 ml of 
hot water as before, centrifuge, and combine the wash water with the other 
centrate. Save this solution for (D) and the residue for (). 

(D) To the centrate from (C) which contains PbCl 2 , add 2 drops of 
NH 4 Ac and 4 drops of K 2 CrO 4 . A yellow precipitate is PbCrO 4 and 
confirms lead. 

(E) The residue from (C) is mixed with 6 drops of dilute NH 4 OH, 
stirred, and 4 drops of H 2 O are added. If all the solid dissolves and no 
dark particles are to be seen, the solid was only AgCl. If the solid turns 
from grey to black, some Hg 2 Cl 2 was present in it and AgCl may have 
been present also and will have to be tested for in the solution. If only 
AgCl is suspected, proceed to (F). If a black precipitate is in evidence, 
centrifuge, collect the centrate and combine it with 10 drops of the 1 M 


NH 4 OH wash liquid used on the mercury residue, and use it as directed in 
(F). Save the black residue for (G). 

(F) The combined centrate and washings from (E) may contain Ag as 
[Ag(NH 3 ) 2 ]Cl. Add a drop of phenolphthalein, then 6 M HNO 3 drop- 
wise until the indicator becomes colorless, then 2 drops more. A white 
precipitate of AgCl separates if silver was present, and it gradually turns 
dark violet in the sunlight, confirming silver. 

(G) The dark residue from (E) is a mixture of Hg(NH 2 )Cl and Hg and is 
a good enough indication by itself to say that the unknown contained 
mercurous mercury. 

If confirmation is desired, it is done according to test 15-4, starting at 
the sign *. 

If the black residue is large and one is working with samples which can 
contain a comparatively small quantity of silver, the mercury residue 
should be tested for silver if none was found above in (F).* This is done 
by dissolving the combined residue in aqua regia, as in test 15-4, which 
solubilizes the metals as chloro complexes, H 2 AgQ 3 , H 3 AgCl 4 , H 2 HgCl 4 , 
etc. If this solution is diluted, AgCl precipitates. It can then be removed 
by centrifugation, washed, dissolved with NH 4 OH, and tested as in (F). 
The diluted aqua regia-mercuric solution may be analyzed as in test 15-9. 


1. Which group 1 metal(s) givc(s): 

(a) A chloride soluble in hot H 2 O? 

(b) A fluoride soluble in H 2 O? 

(c) A yellow phosphate? 

(d) A dark red chroma te? 

(e) A chloride that does not turn dark or dissolve in NH 4 OH? 

2. What single reagent, observation, or test could one use to differentiate 
between the members of each pair: 

(a) Hg 2 Cl 2 and HgCl 2 ? (/) Pb(OH) 2 and Ag(OH) ? 

(b) AgCl and Hg 2 Cl 2 ? (g) AgNO 3 and AgCl? 

(c) Hg 2 Cl 2 and Hg(NH 2 )Cl? (h) HC1 and HNO 3 ? 

(d) PbCl 2 and PbI 2 ? (/) K 2 CrO 4 and K 2 Cr 2 O 7 ? 

(e) HgQ 4 2 and Hg 2 Cl 2 ? (j) MgSO 4 and MgCl 2 ? 

3. Jackson P. Slipshod, a man noted for the rapid formulation of schemes of 

* This is because metallic mercury in step (E) can reduce some silver: 

2Hg f 2Ag(NH 3 ) f 2C1- - 2Ag f 4NH 3 -f Hg 2 Cl 2 , 
and traces of Ag+ would go undetected. 


analyses that are not always rigorous, gives the following three-step method for 
handling a group 1 unknown: 

(a) "Add an iron nail to a portion of the unknown; a silvery deposit indicates 
Ag by virtue of its position in the activity series with respect to Fe. 

(b) "Add PO 4 ~ 3 to another portion; a white precipitate is lead phosphate 
since the solubility of that compound is known to be small. 

(c) "To a third portion add I~ to throw out yellow PbI 2 , a better test than 
using Cl~, since the iodide is much less soluble." 

"With these tests one needs no procedure." 
Are there any pitfalls in the Slipshod technique? 

4. Show by calculation that in test 16-1, 1 drop of 6 M HC1 gives enough Cl~ 
to precipitate all the Ag 1 contained in 2 ml of silver solution containing 10 mg 

5. Balance, showing methods: 

(a) AgCI + Hg = HgCl 2 + Ag 

(/>) AgCI + Zn = Ag + Zn+ 2 + Cl 
(c) Hg 2 Cl 2 + Sir* 8 + Cl'= ? 
(</) Hg./ 2 + NO :i + H 2 O = Hg 2 (OH)N0 3 + H :1 O 
(e) Pb+ 2 + OH = HPb0 2 ~ + H 2 O 

(/') Rewrite equations 4 and 7 from the list of Analytical Reactions of group 1 
cations using NH 3 in place of NH 4 OH. (See p. 242.) 

6. (a) An industrial concern calls you in to give them a test to tell when all 
the brine (NaCl solution) has been rinsed out of a water demineralizer. What 
is your test ? 

(b) There is a question about a certain synthetic detergent as to whether or 
not it contains substantial quantities of Na 2 SO 4 as a cheap process by-product 
(which does not improve or hinder detergency but does add bulk). Assuming 
the remainder of the material is the soluble sodium salt of an organic derivative 
which will not interfere with ordinary analysis, how will you make the deter- 
mination? How could it be done quantitatively? 

(c) A mining company has a crushed ore that assays 20 Ibs of horn silver to 
the ton. The gangue is sand. This ore is good enough to develop if a cheap 
recovery of silver is available. You are asked to propose a method of producing 
silver metal from the ore. 

7. Define: (a) centrate (b) confirmatory test (c) supernatant liquid (d) dis- 
proportionation, and (e) sensitivity of an analytical test. 

8. A colorless solution gives a white precipitate with HC1. Part of the solid 
is soluble upon shaking with a large volume of water and the remainder is 
soluble in strong NH 4 OH. What can one deduce? 

9. One has a solid mixture known to be equal parts of silver metal powder, 
AgCI and PbCl 2 . Organize a flow sheet to effect its separation. 

10. Experiments have shown that the solubility of AgCI at 21 C in pure H 2 O 
is 6.2 mg/liter; in 1 % HC1 is 0.2 mg/liter and in 10% HC1 is 74 mg/liter. Draw 
a rough graph of AgCI solubility vs. % HC1 to illustrate the data, and explain 
the experimental observations in a well organized paragraph. 

11. In the reaction between NH 4 OH and Hg 2 Cl 2 , it is found that half the 



mercury is converted to Hg 11 and half to Hg. Could the reaction have given 
40% of one and 60% of the other, for example, instead? 

12. From reference to a table of solubilities and the preliminary experiments, 
devise an analytical procedure of your own for the separation of the group 1 
ions and their confirmation, assuming only that they are present in solution. 
Explain with the aid of a flow sheet or outline. 

13. In an experiment to determine whether mercurous ion is Hg^ or Hg^ 2 , 
mercuric nitrate was dissolved in water to make three solutions of various 
[Hg+ 2 ]. To each was added a substantial excess of mercury metal and after 
long shaking and equilibrium was established, the concentrations of remaining 
mercuric ion and of mercurous ion formed by the oxidation-reduction were 
determined. From these, equilibrium constants were calculated. The two 
possible reactions and equilibrium statements are: 


Hg+ 2 + Hg = Hg+ 2 

*i = [Hg 2 + 2 ]/[Hg+ 2 ] 

Hg+ 2 + Hg = 2Hg^ 

* 2 = [Hg+] 2 /[Hg+ 2 ] 

Concentrations of ions in the three experiments were found to be as follows: 


[Hg+ 2 ] 



1.2 x 10~ 3 

1.4 x 10- 1 


5.0 x 10~ 3 

5.6 X 10- 1 


8.7 x 10- 3 

10.4 x 10- 1 

Explain what has been proved, and demonstrate with calculations. How 
do you suggest the ionic concentrations be quantitatively determined in this 

Is the value of K found by this experiment consistent with one found from 
cell data? Demonstrate. 

14. When either SOi~ 2 or CrO^ 2 is added to saturated PbCl 2 solution, a 
precipitate forms. If CrO^ 2 is added to saturated PbSO 4 solution a precipitate 
forms. If solid PbSO 4 is heated with NH 4 Ac solution it dissolves, but if PbCrO 4 
instead of PbSO 4 is used, no change is visible. Are these experimental observa- 
tions compatible with one another? How do they establish the relative 
solubilities of the three lead compounds? 


1. J. C. Bailer, J. Chem. Educ., 21, 523 (1944). (Silver oxidation states) 



The general chemistry of lead and mercury was discussed in the last 
chapter. For reasons noted there these elements are in both analytical 
groups. The periodic relationships among the metals of group 2 are 
indicated below. 
































FIG. 16-1. The periodic table in the vicinity of the group 2 metals. 


Bismuth is a hard, brittle metal, sp. gr. 9.80, usually found in the native 
state. Melting at 271 C it forms the basis for low-melting alloys for 




automatic sprinklers, metal baths for heat treating other metals, etc. The 
oxidation states are ( III) as BiH 3 , (III) as BiCl 3 , (IV) as BiF 4> and (V) 
as NaBiO 3 . 

A few compounds are used in medicines but the soluble compounds are 
poisonous and uses for Bi are few. Most Bi salts are insoluble in water 
and the soluble ones hydrolyze readily into complex hydrated oxides. For 
the sake of simplification the bismuthyl ion, BiO 4 is used to illustrate 
hydrolysis but X-ray diffraction work on the solid, usually represented as 
BiOCl, has not shown evidence of this ion. 

H 2 O + BiCl 3 = BiOCl + 2H ( + 2Q- 

This hydrolysis is reversible in the presence of additional HC1. The partial 
Bi + H 2 O = BiO+ + 2H } + 3e~ = - 0.32 

shows that the bismuth ion is about as easily reduced to the metal in acid 
solution as is Cu+ 2 . The reduction is also easily accomplished in basic 
solution, so that Bi 43 , BiO + , or Bi(OH) 3 all go to black Bi under the 
reducing influence of an agent such as stannite ion, Sn(OH)^ 2 . The 
reaction is used as a reliable qualitative test. Bi 2 O 3 in strong alkaline 
oxidizing solutions yields ill-defined hydroxy compounds. We will 
consider the important ion present there is bismuthate, BiO 3 . Bi 2 O 3 
itself is trimorphic, and its molecular weight is variable due to indefinite 
composition. Since Bi(OH) 3 is not amphoteric and Bi 111 does not form 
ammonia complexes, bismuth hydroxide is not soluble in bases. Only a 
few Bi complexes are known, Bil^ is one whose yellow color is sometimes 
utilized in qualitative testing. In acid solution bismuthic acid is an 
oxidizer powerful enough to oxidize manganous ion to permanganate: 

BiO+ + 2H 2 O = 3H<- + HBiO 3 + 2e~ = - 1.6 

Bi 2 S 3 is quite insoluble in water, NaOH, or alkaline sulfides, but will 
dissolve in hot dilute HC1. With CrO^ 2 , (BiO) 2 CrO 4 (?), bismuthyl 
chromate, forms from bismuth solutions and is insoluble in NaOH in 
contrast to PbCrO 4 , which is also yellow. 


Reactions (in order of decreasing [Bi +3 ]) 


Formula Weight 

BiOCl = BiO+ + Cl" 
BiOOH = BiO* + OH~ 
Bi 2 S. a = 2Bi +s + 3S~ 2 

1 x 10~ 9 
~ 1 x 10 12 
1 x 10- 70 



Quantitatively, Bi is determined by precipitation of the hydroxide and 
ignition to and weighing as the trioxide. Colorimetry with thiourea, 
giving yellow complexes such as Bi[CS(NH 2 ) 2 ]+ 3 , is also used. 


This is a very malleable and ductile metal with excellent heat and 
electric conduction properties. It is found native as well as in sulfide and 
Dxide ores; typical minerals are chalcopyrite, CuFeS 2 , azurite, 
2CuCO 3 -Cu(OH) 2 , and malachite, CuCO 3 -Cu(OH) 2 . Ores of this type 
are roasted to produce matte copper, a mixture containing about 50% Cu 
with Fe, S, and SiO 2 . The matte is heated with lime to give an iron- 
calcium-silicate slag and molten copper which is cast into bars known as 
blister copper. These are made the anodes in an electrolysis of a bath of 
CuSO 4 . The cathodes are pure copper rods on which 99.9+ % pure 
copper is deposited. The metal has a sp. gr. of 8.92 and melts at 1063 C. 
It is used in wires, pipes, switches, cables, heat exchangers, etc. In sucfi 
service it may become coated with a protective oxide or sulfide film. 
Alloyed with Zn, Sn, Pb, Ni, etc., Cu makes such important alloys as 
brass, bronze, monel, coinage metal, German silver, and others. Copper 
compounds are used in certain paints, fungicides, and algaecides. 

Dilute or concentrated HNO 3 or HNO 2 dissolves Cu metal, giving 
nitrogen oxides and cupric salts, but, in the absence of oxygen, other dilute 
acids have little effect on the metal. Copper, like silver and gold (which 
comprise group IB in the periodic table), has oxidation states (I), (II), and 
(III), but only the first two are important and readily interconvertible, 
and only Cu +2 will be considered in the analysis scheme later described in 
this chapter. 

Cu+ = Cu+ 2 + e~ E = - 0.15 

Cu(III) can be prepared from Cu(II) in alkaline solution by ozone oxida- 
tion, but it is unstable. The electron structure of the copper atom is 
Is 2 , 2s 2 2/? 6 , 3s 2 3/? 6 3</ 10 , 4s 1 . The 4s electron is readily removed and 
apparently the 3d group, despite being filled, is not too stable, and as many 
as two electrons can be removed from it, giving the higher valence states. 
A few compounds of Cu+ 2 (cupric copper) are water insoluble, as shown 
in Table 16-3, but most are more soluble than corresponding members of 
the cuprous (Cu f ) series. Anhydrous cupric salts are white but in water 
solution are blue due to hydration of the copper ion as Cu(OH 2 )+ 2 . Cu+ 2 
is capable of hybridizing orbitals of the dsp 2 type, which leads to a square 
planar configuration of the four coordinated groups. Cuprous salts are 
generally lightly colored and like Ag f , Cu^ coordinates two groups 



Cupric ion gives a bright blue-green flame test and a blue borax or 
microcosmic salt bead. (See special experiments 4 and 5.) Cu+ 2 is 
easily reduced all the way to Cu, as little as 10 ppm copper giving a visible 

Cu = Cu+ 2 + 2e~ E = - 0.34 


Reactions (in order of decreasing [Cu+]) 


Formula Weight 

CuCl = Cu+ + Cl~ 

3.2 x 10~ 7 


CuBr = Cu+ + Br~ 

6 x 10- 9 


Cul = Cu+ + I- 

1.1 x 10~ 12 


CuSCN = Cu+ -f SCN- 

4 x 10~ 14 


Cu(CN)i~ = Cu+ + 2CN- 

1 x 10- 18 


Cu 2 S = 2Cu+ + S~ 2 

1.2 x 10- 49 


Cu(CN)a = Cu+ -f 3CN- 

5 x 10- 28 



Reactions (in order of decreasing [Cu +2 ]) 


Formula Weight 

CuCO 3 = Cu+ 2 + C<V 

2.5 x 10~ 10 


Cu(C 2 4 ) = Cu^ 2 -h C 2 0r 2 

8.3 x 10~ 12 


Cu(OH) 2 = Cu^ 2 -h 20H- 

1.6 x 10~ 19 


Cu(NH 3 ) 4 + 2 = Cu+ 2 -f 4NH, 

4.7 x 10~ 15 


CuS = Cu+ 2 -h S~ 2 

4 x 10- 36 


The final solution in the group 2A scheme may contain Cu +2 and Cd+ 2 ,* 
and separation of these ions can be accomplished by (a) converting them 
to cyano complexes and then treating with H 2 S, whereupon only CdS is 
capable of precipitating, (6) converting Cu+ 2 to the soluble tartrato complex 
and simultaneously precipitating Cd(OH) 2 in basic solution, or (c) 
selectively reducing Cu+ 2 to Cu without affecting Cd+ 2 by use of Fe powder 
or sodium dithionite. These are illustrated in tests 16-5, 16-7. In each 
of these methods advantage is taken of the reducibility and/or complexibility 
of Cu+ 2 . The reactions are : 

(a) 2Cu(NH 3 )+ 2 + 8CN- = 2Cu(CN)j 2 + (CN\ + 8NH 3 * 

* See equation 23, this chapter, for an alternate reaction. Both are possible depend- 
ing upon conditions. 


(A) Cu+ 2 + 2C 4 H 4 0- 2 = Cu(C 4 H 4 6 ) 2 2 

(c) Cu(NH 3 ) 4 2 + S 2 4 2 + 2H 2 = Cu + 2SO3 2 + 4NH+ 

(Alkaline solutions of Pb TI , Bi lrl , and Sb m are also reduced to the free 
metals by dithionite.) 

Qualitatively, copper is most often recognized by the deep blue color 
of its tetrammine complex, and quantitatively it is determined by electro- 
plating on a weighed platinum electrode or by its action in excess I" 
where it liberates a stoichiometric amount of 1^. The latter is titrated 
with standard sodium thiosulfate: 

2Cu+ 2 + 51- = 2CuI + Ig 
13 + 2S 2 G>3- 2 = 31- + S 4 (V 

The sensitive organic reagent sodium diethyldithiocarbamate (see 
example 7, Chapter 12) is used in quantitative colorimetry for very dilute 
copper solutions. 


This is a silvery, quite ductile, and malleable metal. The chief ore is 
the yellow sulfide, greenockite, CdS, but cadmium is also found in most 
zinc ores, a metal which Cd resembles. TJie ore is roasted at low tempera- 
ture to convert it to sulfate which is acid leached then treated with Zn 
dust to displace Cd and other more noble metals. Cd is volatile and may 
be purified by sublimation to 99.9+ % purity. The metal is used exten- 
sively as a plating for the protection of irons and steels, where it functions 
about as effectively as zinc galvanizing. It is also used in some low- 
melting alloys and as CdS in the more expensive yellow paints. Cadmium 
has been reported to have oxidation numbers of (I) and (II), but evidence 
supporting the lower one is weak and only the latter is considered here. 
The metal's sp. gr. is 8.65 and the m.p. 320.9 C. 

The metal dissolves very slowly in acids giving Cd+ 2 and H 2 . The 
same substances that complex Cu+ 2 usually form the same 4-coordination 
complex types with Cd+ 2 , except that the latter are tetrahedral whereas 
the copper complexes are planar. CdS is the most insoluble of the Cd 
compounds in water and dilute bases, although it will dissolve in dilute 
mineral acids. Cadmium cyanide, phosphate, carbonate, and oxalate are 
water insoluble but will dissolve in dilute acids or NH 3 . Cd is qualita- 
tively identified in group 2, as the only yellow sulfide that is insoluble in 
alkali or alkaline sulfides. 

Quantitatively, Cd is determined, after separation from other metals, 
by precipitation as the carbonate and ignition and weighing as the oxide, 



Reactions (in order of decreasing [Cd+ 2 ]) 


Formula Weight 

CdCl^ = Cd+ 2 + 3C1- 

4 x 10~ 3 


CdC 2 O 4 = Cd+ 2 + C 2 4 2 

1.5 x 10~ 8 


Cd(OH) 2 = Cd+ 2 + 2OH- 

2.0 x 10~ 14 


CdC0 3 = Cd< 2 + CO 3 2 

5.2 x 10~ 12 


CdCNH^ 2 = Cd' 2 + 4NH.J 

1 x 10~ 7 


CdS = Cd< 2 + S- 2 

6 x 10 27 


Cd(CNV 2 = Cd^ 2 + 4CN- 

1.4 x 10~ 19 


or by evaporation of its purified sulfuric acid solution and weighing as 
CdSO 4 . A number of organic reagents are available for Cd analysis; 
see references at the end of Chapter 12. 


Arsenic exists in three (and perhaps four) allotropic forms, but only the 
grey, semimetallic allotrope is stable. Some of its compounds such as 
As 4 O 6 also exhibit polymorphism. The grey arsenic is quite brittle and a 
poor electric conductor. Its sp. gr. is 5.73 and it sublimes at 615 C. The 
common ores are oxides, sulfides, and arsenides like elaudetite, As 4 O 6 , 
orpiment, As 2 S 3 , realgar, As 4 S 4 , and iron arsenide, FeAs 2 . Roasting, 
carbon reduction, and sublimation win the metal from these minerals. 
The element has very limited use in some lead and copper alloys; the 
compounds are all poisonous and are used to kill weeds, insects, and 
rodents; to preserve wood, make war gases, etc. Arsenic has oxidation 
states of (- III) as AsH 3 , (III) as AsCl 3 , (IV) as (As^),, and (V) as 
(As 2 O 5 ) Jc . The last two substances are polymeric and the molecular weights 
are not known. The (IV) oxide may contain both (III) and (V) arsenic, 
since on reaction with base it gives arsenites and arsenates. Due to 
covalent bonding in As 4 S 4 , the oxidation state of As is not (II) as one 
might guess. The metal itself will dissolve in hot concentrated acids to 
give arsenic acid, H 3 AsO 4 , rather than the expected salts. Hydrogen is 
not liberated. Three important partials are: 

As + 2H 2 O = HAsO 2(aq) + 3H+ + 3*r = - 0.25 

HAsO 2 + 2H 2 O = H 3 AsO 4 + 2H+ + 2e~ E = - 0.56 

AsH z = As + 3H+ + 3er E = 0.60 

Heated with oxygen, sulfur, or halogens the metal gives the corresponding 
trivalent compounds. In water solutions As 2 S 3 tends to be colloidal but 


is coagulated by salts. It is soluble in hydroxides and alkaline sulfides, 
its amphoteric character putting arsenic with antimony and tin in cation 
group 2B. Like other members of group VA of the periodic table, it 
forms a hydride, AsH 3 , arsine, when an arsenide is reacted with dilute 
acids or an arsenic(III) compound is reduced in acid solution with an 
active metal such as zinc. Arsine is a poisonous gas but is valuable as a 
means of qualitative testing, giving a dark residue of arsenic when burned 
(Marsh test), or a yellow color with a spot of 50% AgNO 3 (Gutzeit test). 

Four main series of arsenic-containing compounds are common: 
arsenous, As 711 , arsenic, As v , arsenites, AsO 3 3 , and arsenates, AsO 4 3 . 
The latter two are distinguished by addition of A^ 1 , giving a yellow silver 
arsenite and brown silver arsenate. The As 111 compounds are more 
numerous than those of As v ; AsF 5 for example is the only known penta- 
halide of arsenic. 

Very few equilibrium constants involving As are known. 

Arsenous sulfide is soluble in hydroxides or basic sulfides yielding 
arsenites and thioarsenites: 

As 2 S 3 + 4OH- = AsO 2 + AsS 3 3 + 2H 2 O 
and is reprecipitated if the solution is acidified and H 2 S added: 
AsO^ + 2W + 3H 2 S = As 2 S 3 + 4H 2 O 

Precipitation of arsenic(V) sulfide from arsenate solutions is usually 
incomplete because not enough time is given for the slow reaction: 

5H 2 S + 2H 3 AsO 4 = As 2 S 5 + 8H 2 O 

A common procedure is to add NH 4 I to the mixture so that iodide will 
reduce As v to As 111 and quantitative precipitation of As 2 S 3 follows. 

Quantitatively, arsenic may be determined by distillation as AsCl 3 from 
strong HC1 solution. This is followed by titration (oxidation) in basic 
solution with standardized iodine, the end point indicated with starch. 
The main reaction is 

2HCO 3 + H 3 AsO 3 + Ij = H 3 AsO 4 + H 2 O + 2CO 2 + 31 

For small amounts, the Gutzeit test (p. 263) is adaptable to quantity 


This is a silvery, brittle metal; its sp. gr. is 6.68 and m.p. is 630.5 C. The 
chief ore is stibnite, Sb 2 S 3 , which usually occurs black rather than the 
familiar orange sulfide from group 2B. Roasted, the sulfide converts to 
the trioxide Sb 4 O 6 which is separated by sublimation and reduced by 
heating with carbon. The metal finds use in a few low-melting soft 


alloys for bearings, type metal, and storage battery plates. A limited 
number of compounds such as "tartar emetic," KSbOC 4 H 4 O 6 (potassium 
antimonyl tartrate), are used in medicine but most are poisonous as is 
typical of all heavy metals. 

The oxidation states of antimony are ( III), (HI), (IV), and (V), 
representative compounds being SbH 3 , Sb 4 O 6 , (SbS 2 ) x , and KSb(OH) 6 . 
There are no well defined Sb lv compounds and some of the others are also 
polymeric and of uncertain molecular weights. Acids usually convert the 
metal to Sb 4 O 6 , strong HC1 giving antimony trichloride from that. SbCl 3 
and similar compounds hydrolyze strongly to give insoluble oxy (or 
suboxide) derivatives, as SbOCl, whose formulas are not definite and are 
written only as approximations in equations. The antimonyl ion, SbO+, 
has not been shown present in these materials which are polymers of 
covalently bonded Sb and O atoms. The trioxide Sb 4 O 6 is amphoteric 
and will dissolve in bases to give slightly soluble (meta) antimonites as 
NaSbO 2 -2H 2 O which are powerful reducing agents. Hot concentrated 
HNO 3 on Sb 4 O 6 yields some (ortho) antimonic acid which, unlike the 
corresponding arsenic acid, has the formula HSb(OH) 6 in which the metal 
coordination is 6. The sulfides of antimony(III) and (V) are soluble in 
alkaline sulfide solutions, respectively forming thioantimonites, SbS;r, and 
thioantimonates, SbS 4 3 . This behavior puts Sb with As and Sn in group 2B. 

Potentials related to the discussion are: 

x = Sb + 3H^ + 3<r = 0.51 

showing the ease of oxidation of the ( III) state, and its acidic character; 

Sb + H 2 = SbO+ + 2H+ + 3r = - 0.21 


SbO+ + 5H 2 = HSb(OH) 6 + 3H+ + 2e~ E ^ - 0.7 

showing the tendency of Sb 111 and Sb v to be reduced in acid solution, 
whereas in alkaline solution, oxidation is easy: 

2S~ 2 + SbS- = SbS 4 3 + ler E^ = 0.6 

Compare these with the arsenic equations in the preceding section. 

Equilibrium constants are not accurately known for most Sb reactions. 

The brilliant orange color of Sb 2 S 3 and the evolution of stibine gas, 
SbH 3 (which is tested like AsH 3 ), are two good qualitative indications of 

Differentiation between AsH 3 and SbH 3 can be made in several ways. 
With the Gutzeit test, both give a yellow or yellow and black color on the 
AgNO 3 spot, but, by using NaOH instead of H 2 SO 4 in the generator (the 
Fleitmann modification), only AsH 3 is released. If the Marsh test is 


used, hypochlorite or hypobromite will dissolve the deposit of metallic 
arsenic but not antimony, because of the former's greater ease of oxidation : 

2As + 50C1- + 6OH- = 2AsO 4 3 + 5C1- + 3H 2 O 

Quantitatively, Sb is analyzed by an iodometric method; it is com- 
plexed with tartrate which not only keeps Sb 111 in solution, but also 
lowers the oxidation potential so that triiodide quantitatively oxidizes 
ittoSb v : 

SbOC 4 H 4 Oe + 17 + H 2 = SbO 2 C 4 H 4 O^ + 3I~ + 2H+ 
Starch is the indicator in iodine titrations. 


This metal has three allotropic forms, the a or white form being the 
common, stable one at room temperature. Its sp. gr. is 7.31 and its 
m.p. is 231.8C. At low temperatures this is reversibly convertible to 
gray or /3-tin. The gray form is brittle and easily powdered, whereas the 
white form is ductile and malleable unless heated to temperatures above 
232 C where transformation to y-tin, another brittle allotrope, occurs. 
The chief ore is cassiterite, SnO 2 , which is reduced by carbon; the metal is 
electrolytically refined. Tin is used in electrolytically-prepared tin plate 
for "tin" cans and in some important soft alloys like bronze, babbit, 
solder, pewter, type metal, and pot mptal. 

Tin has three oxidation states, (II), (III), and (IV), and being amphoteric 
forms four main compound types : stannous, Sn IJ ; stannic, Sn lv ; stannites, 
SnOj (or Sn(OH)3); and stannates, SnOj 2 (or Sn(OH)j 2 ). Little is 
known about tin(III). The metal dissolves in HC1 giving H 2 and the 
tetrachloro complex SnQ 4 2 , but with HNO 3 , metastannic acid, H 2 SnO :j 
(or hydrated stannous oxide, SnO 2 (H 2 O)J is the result. Salts of both 
tin valences are considerably covalent, ionize feebly, hydrolyze readily, 
and must be kept in acid solution or precipitation of hydrated oxides 
occurs. The sulfides are soluble in acid, base, or alkaline sulfides and 
polysulfides, the last reagents giving thiostannate, SnSg 2 , from either 
brown SnS and yellow SnS 2 . 

Three tin half reactions are 

2e- E = 0.14 

Sn+ 2 = Sn lv + 2e~ E = - 0.15 

SnS + 2S- 2 = SnSg 2 + 2e~ E K = > 0.6 

Equilibrium constants are not known accurately for most tin reactions. 
The K SP of Sn(OH) 2 = 3 x 10' 27 , of Sn(OH) 4 c* 10~ 57 and of SnS 
= 1 x 10- 26 . 


In the analytical process for group 2B, Sn lv and Sb m come out in 
solution together and are tested in each other's presence. This is done by 
(a) complexing tin with oxalate and precipitating antimony with H 2 S 
and/or (b) reducing antimony to black Sb and tin to Sn 11 with iron arid 
acid, then reacting the latter with mercuric chloride to effect reduction to 
white Hg 2 Cl 2 and black Hg. The tin reactions in these steps are 

(a) Sn lv + 3C 2 4 2 = Sn(C 2 O 4 )3 2 
Sn(C 2 O 4 ) 3 2 + H 2 S = No reaction 

(b) Sn TV + Fe = Fe+ 2 + Sn 11 

2Sn ir + 4HgCl 2 = Hg 2 Cl 2 + 2Hg + SnCl^ 2 

The stannous solution can also be tested for Sn ir by means of a flame test; 
the color is bright blue. 

Quantitative methods usually call for precipitation of H 2 SnO ;{ from 
HNO 3 solution, then filtration on ashless paper, and ignition at 800 C to 
SnO 2 . If the original precipitate is impure and the SnO 2 colored, tin can 
be volatilized from it as SnI 4 by heating with NH 4 I, and the tin content 
calculated by the loss in weight. 

Group 2 Analysis General Description 

A solution containing only group 2 ions, or a general unknown solution 
that has been freed of group 1 ions except some Pb* 2 , is treated with 
NH 4 OH and HC1 to adjust the p\i at about 0.6. The mixture is then 
treated directly with H 2 S gas or with thioacetamide which hydrolyzes to 
release H 2 S, and precipitation of the group 2 sulfides takes place.* If 
thioacetamide is used, the mixture is then diluted to give a pH of about 
1 .0 and reheated to complete the precipitation reaction. By this procedure 
the Xtf/.'s of group 2 sulfides alone are exceeded and group 3, 4, and 5 
metals remain in the centrate when the mixture is separated by 

The group 2 residue is treated with hot, 6 M KOH and centrifuged, 
dividing it into those elements whose sulfides are not amphoteric and 
hence not soluble [HgS (some), PbS, Bi 2 S 3 , CuS, CdS] and those whose 
sulfides are amphoteric and base-soluble [HgS (some), As 2 S 3 , Sb 2 S 3 , 
SnS]. The residue is called subgroup 2A or the "copper group," the centrate 
is subgroup 2#, or the "tin group." 

The group 2A residue is stirred with dilute HNO 3 , putting into solution 
all the sulfides except HgS, which remains after centrifuging as HgS or 
as the insoluble double salt 2HgS-Hg(NO 3 ) 2 . This is dissolved in aqua 

* Strong oxidizing agents like HNO 3 must be absent or H 2 S is oxidized. In general 
these substances are not present or are reduced before group 2 is begun. 


regia and confirmed with Sn 11 as in group 1. The centrate from HgS 
removal is mixed with dilute H 2 SO 4 , evaporated to drive off HNO 3 
(which interferes with the formation of PbSO 4 ), then diluted and centri- 
fuged, leaving PbSO 4 as a residue if Pb+ 2 is present, and retaining Bi m , 
Cu +2 , and Cd+ 2 in solution. Lead is confirmed by soli ion of PbSO, 
in NH 4 Ac and reprecipitation as yellow PbCrO 4 . Addition to the 
Bi Cu Cd mixture of excess NH 4 OH complexes copper and cadmium 
as soluble tetrammine ions, blue Cu(NH 3 )+ 2 and colorless Cd(NH 3 )J 2 , 
and precipitates any Bi 111 present as white Bi(OH) 3 (or BiO(OH),,). This 
residue when treated with stannite ion, Sn(OH)~, yields a confirmatory, 
black residue of metallic bismuth. If the centrate from the bismuth 
separation is dark blue, copper is present but Cd 11 , being colorless, must be 
tested further. Thioacetamide or H 2 S is added to the mixture of com- 
plexes, a black precipitate showing CuS and a yellow precipitate showing 
CdS if those metals are present. If copper was present, CuS would 
generally obscure any yellow CdS so NaCN is next added. The black 
sulfide will dissolve to form the very stable Cu(CN)~ 2 , leaving the cadmium 
precipitate unaffected. The yellow color confirms Cd f2 . (Two alternate, 
noncyanide methods are also given in the formal procedure.) 

The group 2B centrate described in the second paragraph above, is 
mixed with dilute HC1 and centrifuged. The residue contains some S and 
HgS and all the As 2 S 3 , Sb 2 S 3 , and SnS. Hot dilute HC1 dissolves the 
latter two sulfides and they are separated with the centrifuge. The 
residue of the other two is reacted with a mixture of dilute NH 4 OH and 
H 2 O 2 , dissolving As 2 S 3 , forming AsO 4 3 , which later is confirmed by 
precipitation as white (NH 4 )Mg(AsO 4 ), and leaving black HgS which may 
be combined with that found in group 2A or tested separately as previ- 
ously explained. Generally most of the HgS is isolated in group 2A. 
The HC1 centrate of antimony and tin contains such chlorocomplexes as 
SbCl 4 and SnCl~ 2 , and is divided between two tubes. To one is added 
oxalic acid which complexes tin from further action, and, when H 2 S (or 
thioacetamide plus heat) is added, orange Sb 2 S 3 comes down and the color 
is sufficient to prove the ion. Iron filings are used to reduce tin in the other 
tube to Sn 11 , and it may be tested with Hg 12 ; a white-to-grey precipitate 
of Hg 2 Cl 2 + Hg is evidence of reaction as in group 1. Antimony(III) 
is reduced to Sb by the iron and does not interfere. 

Other tests for group 2 ions such as the Gutzeit, flames, and organic 
reagents are listed in the preliminary studies of these ions and in the 
procedure for unknowns in later paragraphs of this chapter. A flow 
chart of the processes just described is given on pp. 258, 259. 

See also part 4 of special experiment 9 for group 2 analysis by paper 




2A 2B 


Residue Residue Centrate 





AsS 2 

to be 

Bi 2 S 3 , 

HgS 1 


Solution m<1 




KOH, heat . 

SbS 2 

except > 

i H 2 S, 
group 1 centrifuge 

later for 

+ CdS 7 
As 2 S 3y 

- + Sn(OH), 2 
SnS 3 2 

centrifuge ^ o 


Sb,S 30f 


HgS 2 2 

SnS Br j 

HNO a , S 2* 

heat, " etc. 




Centrate Residue 

PbCr0 4 


then J . , 

Sn( H) '~ 

NH 4 OH, 


Cu(NH 3 ) 
Cd(NH 3 )+ 2 , 


2HgS-Hg(N0 3 ) 2l , 

HN0 3 , 
then SnCl, 2 

Hg 2 CI 2|y 


CdS r 

Cu(CN) 3 2 

H 2 S Gas as a Precipitation Agent 

Hydrogen sulfide gas is toxic and should always be handled with caution 
in good ventilation. The first signs of poisoning are nausea and headache 
and must be immediately relieved with smelling salts and/or copious 
breathing of fresh air. In larger doses H 2 S is lethal. 

H 2 S may be supplied in the laboratory in a number of ways. A 
common source is ferrous sulfide, which is reacted with acid in a student 
generator or some modification of a central Kipp generator: 

FeS + 2H+ = Fe+ 2 + H 2 S 

Another gas-releasing reaction is that between sulfur and paraffin wax 
initiated by heating, and approximately* represented as 

C3oH 62 + 31S = 30C + 3\H 2 S 
Because H 2 S is poisonous, corrosive, and ill-smelling, it is always best to 

* Paraffin is a mixture of alkane hydrocarbons from about C 22 to C 30 . The com- 
position of the residue is more complex than represented. 




FLOW SHEETGROUP 2 continued 
Centrate Residue Centrate Residue 


2 S 3 

NH 4 OH, 

H 2 2 , 


AsO- 3 + 

NH 4 OH, 
NH 4 ', 


s ! 


(combine \ 
in group 
2A / 

NH 4 MgAsO 4H , 

generate or use only the minimum quantity which will do the job. Air 
pollution with H 2 S should be avoided and reactions and evaporations 
should be conducted under the hood if possible. 

Because of the inconveniences, a number of non-H 2 S systems of qualita- 
tive testing have been invented. For instance, Carnog bases his on the use 
of alkaline solutions and (NH 4 ) 2 S; West, Vick, and LeRosen have devised 
a system utilizing a benzoic acid-sodium benzoate buffer; and Dobbins 
and Gilreath suggest a sequence with HC1, NH 4 (H 2 PO 4 ), and pyridine- 
SCN- (see references, this chapter). As yet, however, the original 
hydrogen sulfide system of Fresenius (1850) is still the one most widely 
taught. The quantitative separations it affords plus the well investigated 
yet diverse reactions, theories, tabular data, problem possibilities, and 
practical applications are features in its favor. 

Thioacetamide as a Source of H 2 S 

For the reasons just mentioned the H 2 S system is retained in this edition, 
but an alternate source of H 2 S for the operator's convenience and comfort 


seems desirable. In 1949 Barber and Grzeskowiak suggested the use of 
thioacetamide as a source of H 2 S for qualitative analysis. The compound 
was prepared originally in 1877 by Hofmann and Bernthsen and its 
synthesis is detailed in special experiment 12 with references to original 
literature. Thioacetamide hydrolyzes to slowly release H 2 S in hot, dilute 
strong acids (group 2 precipitating conditions) or in hot, dilute strong 
bases, but is quite stable at room conditions in aqueous solution at/?H 7. 
This obviates the use of gas generators, gives better H 2 S control, denser 
precipitates, and generally is a neater and handier technique. 

Thioacetamide is a white, crystalline material, somewhat soluble in 
water and benzene. The chemical is usually used in the form of its 
28% aqueous solution. At least 5 minutes should be allowed for its 
reaction with strong acid or base at hot-water bath temperatures. 
Reactions are: 

(pH < 7) CH 3 Cf + 2H 2 + H 

rapid \NH 2 

4 CH 3 Of + NHt 4- H 2 S 

= 7) CH 3 C<( + H 2 

very slow 

= CH 3 -C 

(ptt > 7) CH 3 C< + 30H- 

rapid \NH 2 

NH 2 

= CH C + Atf/ 3 + H 2 

3 3 2 

As directed by the instructor, either H 2 S gas or thioacetamide may be 
used with this text for the cation group 2 analysis. Directions are given 
for the organic reagent. In every case where 8% thioacetamide solution 
is used it is to be understood that saturation with H 2 S gas is an equivalent 


In all the preliminary tests, use test solutions from the side shelf which 
are made for that purpose and contain 10 mg of the metal ion per milliliter 
of solution. A few tests are to be made on more concentrated solutions 


and are specially noted. Most of the reactions may be carried out in 
3- or 4-in. test tubes. The thioacetamide solution in each case is an 8% 
aqueous solution. Questions asked in the text as well as volunteer 
observations are to be answered in the laboratory notebook. 

Test 16-1. Sulfides of the Group. Label nine test tubes as Hg+ 2 , Pb+ 2 , 
Bi IIT , Cu+ 2 , Cd+ 2 , As m , Sb lir , Sn 11 , and Sn lv . Put 5 drops of each single soln. 
in its tube plus 10 drops of water. To each add 8 drops of 8% thioacetamide 
and to the Hg+ 2 , Pb+ 2 , Cu+ 2 , and Cd 42 tubes add a drop of 6 M HCI (the other 
solns. are already acidic). Heat the tubes in a gently boiling water bath for 
5 min, then add 10 drops of water to each, and reheat. Note the pptn. order 
and sulfide colors of the ions of group 2. Centrifuge, or allow the ppts. to 
settle, and decant, discarding the supernatant liq. in either case. Add 12 drops 
of H 2 O, 1 drop of thioacetamide, and 4 drops of 6 M KOH to each tube, and 
replace them in the hot water bath. During the next 5 min intermittently stir 
each and note the results. The group 2B sulfides are alkali-soluble, the 2A 
sulfides are not, and the subgroups are therefore separable. In tabular form, 
give all results in your notebook including colors and formulas of substances 
observed. (Reaction equations are given in the section following these pre- 
liminary tests.) 

Test 16-2. Some Iodide and Phosphate Reactions; Possible 
Confirmatory Methods (Optional), (a) Try a few drops of each group 2 
ion with a drop of I~ test soln. Then add several more drops of I~ to each, 
with mixing. The more distinctive reactions which may aid one in spot testing 
are: (1) red HgI 2 dissolving to form yellow Hgl^ 2 (2) orange Bi complexes like 
Bil t ^ 2 , Biljr 3 , etc., and (3) reduction of Cu+ 2 to Cu+ and production of brown I 2 
and whitish Cul. Write the ionic reactions and note that if the unknown 
contains certain combinations of cations, this iodide test could be useful in 

(b) Make a soln. of Na 3 PO 4 using about 0.2 g of crystals and 2 ml of H 2 O. 
To several drops of each of the group 2A solns., add 8 drops of phosphate soln. 
and 5 drops of dil. HCI and mix. Note if any phosphates remain undissolved. 

Test 16-3. Hydrolysis Reactions of Bi and Sb. Bi+ 3 , Sb+ 3 , Sn+ 4 , Sn f2 
and As+ 3 hydrolyze (Chapter 10), yielding water-insol. ppts. which can be 
separated, redissolved in HCI and tested further. The fact that the unknown 
contains a ppt. or is strongly acidic may indicate one of these ions. See equa- 
tions 14, 16, 49, this chapter. 

Put 20 ml of H 2 O in a 50-ml beaker and heat to boiling. Slowly add 3-6 
drops of Sb test soln. The ppt. is SbOCl. Write the equation using SbCljf 3 
as a reactant. Why does the reaction shift toward the hydrol. products upon 
diln. ? Why is the shelf soln. of SbCl 3 approx. 3 M in HCI ? 

Test 16-4. The Reinsch Method. This test is used in criminology as 
a screening investigation for heavy metal poisons. To 5 drops of Hg+ 2 test soln., 
add 3 drops of dil. HCI and a small piece of copper sheet, copper penny, or a 


coil of copper wire. Heat in a water bath for several min, then remove the 
copper, wash, and examine its surface. Note the deposit and the way in which 
rubbing shines it. This plating is characteristic of heavy metals, though the 
shiny appearance is typical only of mercury. (Other metals give a flaky or hard 
grey to black ppt.) List other group 2 metals that would give the test. Try 
one or two of them and write equations for the reactions. How can this 
method be modified if only one drop of Hg+ 2 soln. is available for testing? Try 
your idea and report the result. 

Test 16-5. Cd+* in the Presence of Cu^ Cyanide Method. Put 4 

drops of each of the Cd+ 2 and Cu+ 2 test solns. in a tube, add 10 drops of 
H 2 O and 6 drops of dil. NH 4 OH, and shake. The deep blue color is 
proof that Cu +2 is present as the tetrammine complex.* * Add 12 drops of 
thioacetamide and heat the mixt. for several min in the water bath. (Both 
sulfides ppt. since the stability of the NH 3 complexes is not great enough to 
prevent it.) Remove the tube, centrifuge, and discard the centrate. Add a few 
crystals of NaCN (or NaCN soln.) and 10 drops of H 2 O to the residue and stir. 
The Cu 2 S should gradually dissolve, forming the colorless tricyano copper(I) 
ion, leaving yellow CdS as evidence that cadmium is present. (See calcn., 
p. 149, Chapter 9.) 

Test 16-6. Cd^ in the Presence of Cu^ 2 - Dithionite Method. 

To about 1 ml of Cu+ 2 -Cd+ 2 mixt. add 6 drops of dil. NH 4 OH and * heat in a 
water bath. When hot, add 60-80 mg of sodium dithionite, Na 2 S 2 O 4 (also 
called hyposulfite or hydrosulfite) and keep the mixt. hot. Add more dithionite 
after several min. and reheat for 5 min. If the blue color is not entirely dis- 
charged, add more dithionite and reheat. Centrifuge, discard the metallic Cu, 
and to the centrate, add 2 drops more of NH 4 OH, 10 drops of thioacetamide, 
and reheat. A yellow ppt. is CdS. This test works as a spot reaction for Cd +2 
in group 2 unknowns if arsenic is absent. (See reference 3, p. 276.) 

Test 16-7. Cd+ 2 in the Presence of Cu+ 2 . Tartrate Method. *Place 
in a crucible about 1 ml of the Cu+ 2 -Cd f 2 mixt., acidify with dil. HNO 3 , and 
add 2 drops in excess. In the hood, slowly evap. the soln. and bake it for several 
min to destroy NH+ salts (which would be present from the Bi(OH) 3 pptn. in 
general unknown sample procedure). When cool, add 1 drop more of HNO 3 , 
1 ml of H 2 O, then 8 drops of 6 M NaOH, 1 ml of 0.1 M sodium potassium 
tartrate, NaK(C 4 H 4 O 6 ), and mix well. Transfer the mixt. to a test tube and 
centrifuge. The white ppt. is Cd(OH) 2 , the blue soln. is a tartrate-copper 
complex, Cu(C 4 H 4 O 6 )2~ 2 , and this blue color confirms copper. Wash the residue 
twice with H 2 O, discard the washings, suspend the residue in about 1 ml of H 2 O, 
add 1 drop of NaOH and 6-10 drops of thioacetamide, and heat in a water bath. 
Yellow CdS proves Cd+ 2 . (See reference 6, p. 276.) 

* * This sign refers to procedure (J). Ignore the signs when only making preliminary 
tests. If one is working with an unknown sample, however, he may start at these 
points with his Cd-Cu mixture and use the directions for Cd+ 2 analysis. 


Test 16-8. Sodium Diethyldithiocarbamate for Cu+ 2 . Review ex- 
ample 7 of Chapter 12, p. 194. Prepare a series of dilutions contg. 0.02-5 
ppm Cu+ 2 in 4-in. tubes. Add 4 drops of 2 M NH 4 OH and 2 drops of organic 
reagent (Table A-15), and note the colors. Notice also that because the soln. 
is so dil., no blue color of ammine complex is observed when NH 4 OH is added 
yet the organic reagent gives a good test series. The color gradation is used 
in quant, estimation. An unknown sample may be diluted considerably to 
minimize interferences and still yield this test for Cu. 

Test 16-9. The Gutzeit Method for Arsenic and Antimony. Prepare 
the semimicro Gutzeit apparatus as diagrammed in Fig. 14-4. Wet the lower 
half of the paper strip with 50% AgNO 3 . Put about 150 mg of As-free zinc 
granules in the test tube and add 6 drops of H 2 O and 1 drop of dil. H 2 SO 4 . 
Put the top on and note that the AgNO 3 spot on the paper does not turn color 
in the presence of H 2 gas. A "blank" determination like this is always advisable 
to check the purity of the reagents. Remove the top, add 2 drops of As TTt test 
soln. or 6 drops of unknown, put in a loose bit of cotton to prevent spotting of 
the paper by splattering, replace the top, and observe the reaction for the next 
few min. If too much foaming occurs, cool tube under the tap; if reaction is 
too slow add another drop or two of acid or warm the tube. The method can be 
semi-quantitative in the range 0.3-3 ppm of As, if one measures the length of 
the yellow area on the paper. 

(Optional) Saturated HgCl 2 or 20% HgBr 2 in methyl alcohol have been 
suggested in the literature as substitutes for AgNO 3 . The interested student 
might try impregnating filter paper in tjiis way and report on the diff. color, 
sensitivity and reaction rate. 

Interferences to the test include phosphides which liberate PH 3 , antimony 
compounds which give SbH 3 , and sulfides which give H 2 S. These poisonous 
gases react to give colored products with AgNO 3 . 

If only sulfide is an interference it may be absorbed by impregnating the 
cotton with PbAc 2 solution and trapping H 2 S as black PbS. Antimony is an 
interference; see test 16-10. Mercury compds. also interfere but in a different 
way: they ppt. mercury, which amalgamates with the zinc, slowing its reaction. 
Pentavalent arsenic should be reduced with NH 4 I or Sn 11 before running this 
test. Repeat the test using Sb test soln. instead of As. 

Wash the apparatus well after each test. Any left over, solid zinc should go in 
the waste jar, not the sink. 

Test 16-10. The Fleitmann Method for Arsenic. Repeat test 16-9 
using As 111 , 6 drops of 6 M NaOH in place of H 2 SO 4 and a fresh strip of AgNO 3 
spotted paper. The tube will need intermittent heating as the alkali reacts 
slowly with the zinc. Antimony under these conditions does not give SbH 3 
and hence is not an interference as AsH 3 alone is released. 

Test 16-11. Reduction of Sb 111 . A drop of Sb m test soln. plus a drop 
of dil. HC1 are placed on a piece of platinum foil or silver coin, and a small piece 
of tin is placed in contact with one side of the drop. A rapidly forming black 


deposit is free antimony metal. Mercury, bismuth, and arsenic are also dis- 
placed by this procedure. If only arsenic is suspected of being present, it may be 
dissolved in dil. sodium hypochlorite solution, NaOCl (distinction from Sb). 
Verify this and write equations. 

Test 16-12. Rhodamine-B Test for Antimony. Whereas Bi m and 
Hg+ 2 are interferences in cation group 2 for this test, it is valid in the presence of 
Sn TT , which makes it worthwhile, since antimony and tin appear together in the 
last reactions of the group 2B scheme. This is our first test in which a spot 
plate is used. Put several drops of Sb m soln. on a spot plate and acidify with 
2 drops of dil. HC1. Add a crystal of KNO 2 and when reaction is over, and 
antimony is oxidized to Sb v , add 2 drops of the dye soln. A pink color turning 
violet or giving violet flecks is a positive test for Sb. Compare the color with a 
blank and with Sn u and the reagent. (See rhodamine-B, example 9, Chapter 1 2.) 
The violet compd. can be extracted by shaking with a few drops of benzene, 
C 6 H 6 , if this test is run in a test tube. The extraction is an extra 'confirmation 
of Sb. 

Test 16-13. Ammonium Molybdoar senate. In the absence of PO^ 3 , 
and in the presence of HNO 3 and ammonium molybdate, (NH 4 ) 6 Mo 7 O 24 4H 2 O, 
arsenic is converted to arsenate, AsO 4 3 , and reacts in hot soln. to give a yellow 
ppt., H 4 (NH 4 ) 3 [As(Mo 2 O 7 ) 6 ]. See p. 392, on the analogous phosphate re- 

Put 3 drops of As 111 test soln. in a tube, add 3 drops of coned. HNO 3 , and 
warm for 2-3 min. Pour this into a second soln. made by dissolving a crystal 
of ammonium molybdate and several crystals of NH 4 NO 3 in 10-12 drops of 
dil. HNO 3 . Keep the mixt. warm in a bath for several min and note the gradual 
appearance of the yellow ppt. The concentration limit is about 75 ppm As. 

Test 16-14. Flame Test for 5n H . Put 10 drops of Sn IV test soln. in a 
tube and add about 100 mg of powdered iron and 4-6 drops of dil. HC1. Allow 
the reaction to proceed 5 min. to effect reduction of tin to the dipositive state. 
Pour a little of this soln. in the bottom of a crucible lid and let it flow around to 
wet a max. area. Hold the lid, bottom down, in a hot bunsen flame and notice 
a bright blue flame playing over the porcelain surface. The test only works 
with stannous tin and will detect amounts larger than about 2 ppm. 

Test 16-15. The Reducing Action of 5n u . As will be shown in 
cation group 3, Fe+ 3 forms a deep red colored Fe(SCN)+* complex, but Fe +2 
does not. To 1 drop of Fe+ 3 tests soln. add 1 drop of dil. HC1, 10 drops of H 2 O, 
and 2 drops of KSCN soln. Add this dropwise to 6 drops of Sn 11 test soln. 
Write the reaction equations and account for the observed result. Support 
your conclusion with data from the oxidation potentials table. 

Analysis of a Known Mixture 

When the student has finished required preliminary studies and recorded 
the results in his laboratory notebook he may be directed to analyze known 


mixtures of groups 2A, 2B, or a combination of ions of both subgroups. 
If not, he is then ready to run the unknown sample. In either case, one 
portion of sample should be used to run individual tests as given in the 
prelims or suggested by them, another portion used in following the 
formal procedure as it appears in the next section, and another portion 
saved for further tests if results on the first two parts do not appear 

A. Sulfide Precipitations 

^S ,O 

CH 3 C + 2H 2 + H+ = CHzC^ + NH+ 


1. Hg+ 2 + H 2 S = HgS* + 2H+ 

3Hg^ 2 + 2H 2 S + 2C1- = HgCl 2 2HgS, F + 4H+ 

2. Pb 42 + H 2 S = PbS* + 2H+ 

3. 2Bi ul + 3H 2 S = BigS^ + 6H+ 

4. Cu+ 2 + H 2 S = CuS,, + 2H+ 

5. Cd+ 2 + H 2 S = CdS r + 2H+ 

B. Mercury 

6. 3HgS + 8H^ + 12C1- + 2NOg = 3HgCl4 2 + 2NO + 3S r + 4H 2 O 

7. 2HgClj 2 + SnClj 2 = Hg 2 Cl 2w , + SnCl^ 2 + 4C1~ 
Hg 2 Cl 2 + SnCl^- 2 = 2H gll + SnCl' 2 

C. Lead 

8. 3PbS + 8H^- + 2NOg = 3Pb+ 2 + 2NO + 3S + 4H 2 O 

9. Pb+ 2 + SOj 2 = PbS0 4w , 

10. PbS0 4 + 2Ac- = PbAc 2 + SOj 2 

11. PbAc 2 + CrOj 2 = PbCr0 4r + 2Ac~ 

D. Bismuth 

12. Bi 2 S 3 + 8H+ + 2NOg = 2Bi+ 3 + 2NO + 3S + 4H 2 O 

13. 2Bi+ 3 + 6NO- + 6H+ + 3SO 4 2 

= 2Bi+ 3 + 3SO 4 2 + (6H+ + 6NO3) (Boil off) 


14. 2Bi+ 3 + 3SO4 2 + 2H 2 O = (BiO) 2 S0 4jy + 4H+ + 2SO 4 2 

15. 2Bi+ 3 + 6NH 4 OH = 2Bi(OH) 3w , + 6NH+ 

16. Bi(OH) 3 + 3H+ = Bi+ 3 + 3H 2 O 

Bi+ 3 + 3d- + H 2 O = BiOC! w + 2H+ + 2C1~ 

17. Bi(OH) 3 + 3Sn(OH)- 2 + 3OH = 2Bi* + 3Sn(OH)- 2 

. Copper 

18. 3CuS + 8H+ + 2N0 3 = 3Cu +2 + 2NO + 3S + 4H 2 O 

" Hu ** 

19. Cu+ 2 + 2NOg + 2H+ + SOj 2 

= Cu 42 + SO^ 2 + 2H+ + INOj (Boil off) 

20. 2Cu+ 2 + SOj 2 + 2NH 4 OH = Cu 2 (OH) 2 SO 4flu + 2NH 4 ' 
Cu 2 (OH) 2 S0 4 + 8NH 3 = 2Cu(NH 3 )+ 2 u + SOj 2 + 2OH- 

21. Cu(NH 3 )| 2 + 4H+ = Cu+ 2 + 4NH+ 

22. 2Cu+ 2 + Fe(CN)^* = Cu 2 Fe(CN) 6flr 

23. 2Cu(NH 3 )+ 2 + 7CN- + H 2 O 

= 2Cu(CN) 3 2 + CNO- + 6NH 3 + 2NH+ 

24. Cu(NH 3 )| 2 + H 2 S = CuS + 2NH+ + 2NH 3 

25. Cu(CN) 3 2 + H 2 S = N.R. 

26. Cu 2 + Fe = Cu + Fe +2 

27. Cu(NH 3 ) + 2H 2 + S 2 O 4 2 = Cu Br + 2SO, 2 + 4NH+ 

F. Cadmium 

28. 3CdS + 8H+ + 2NO 3 = 3Cd+ 2 + 2NO + 3S + 4H 2 O 

29. Cd 12 + 2NO 3 + 2W + SOj 2 

= Cd+ 2 + SO 4 ' 2 + (2H+ + 2NO 3 ) (Boil off) 

30. Cd+ 2 + 4NH 3 = Cd(NH 3 )+ 2 

31. 2Cd+ 2 + SOj 2 + 2NH 4 OH = Cd 2 (OH) 2 SO 4lt , + 2NHJ 
Cd 2 (OH) 2 SO 4 + 8NH 3 = 2Cd(NH 3 )+ 2 + SO 4 2 + 2OH- 

32. Cd(NH 3 )+ 2 + 4CN- = Cd(CN) 4 2 + 4NH 3 

33. Cd(CN) 4 2 + H 2 S = CdS + 2HCN + 2CN~ 

34. Cd+ 2 + 2OH- + C 4 H 4 Og 2 = Cd(OH) 2w . + C 4 H 4 <V 



G. Sulfide Precipitations 

35. 2As0 3 3 + 6H+ + 3H 2 S = As 2 S 3r + 6H 2 O 

36. 2SbCle 3 + 3H 2 S = Sb 2 S 3(yr + 6H+ + 12C1- 

37. SnClg 2 + 2H 2 S = SnS 2r + 4H* + 6C1~ 
SnClj 2 + H 2 S = SnS B , + 2H+ + 4C1- 

H. Hydroxide Solutions 

38. 2As 2 S 3 + 40H- = AsO 2 + 3AsS 2 + 2H 2 O 

39. 2Sb 2 S 3 + 40H- = Sb(OH)j + 3SbS 2 

40. 3SnS 2 + 60H- = Sn(OH) 6 2 + 2SnS3 2 
2SnS + 30H- = SnS 2 2 + Sn(OH) 3 

I. Acid Reprecipitations 

41. As0 2 + 3AsS 2 + 4H+ = 2As 2 S 3 + 2H 2 O 

42. Sb(OH)j + 3SbS 2 + 4H< = 2Sb 2 S 3 + 4H 2 O 

43. Sn(OH)g 2 + 2SnSg 2 + 6H f = 3SnS 2 + 6H 2 O 
SnS 2 2 + Sn(OH)3 + S 2 2 + 3H+'= 2SnS 2 + 3H 2 O 

J. Arsenic 

44. 2As 2 S 3 + 120H- + 14H 2 O 2 = 2^0? + 20H 2 O + 

45. AsOj 3 + NH+ + Mg+ 2 = NH 4 MgAsO 4if , 

46. Gutzeit test applies also to antimony. 

As 111 + 4Zn + 5H+ = AsH a + 4Zn+ 2 + H 2 
AsH 3 + 6Ag+ + 3NOj = Ag 3 As-3AgNO 3ir + 3H+ 
2H 2 = 6Ag fl + HAsO 2 + 3H< 

K. Antimony 

47. 2SbCV + 3Sn ^ 

48. Sb 2 S 3 + 6H+ + 12C1- = 2SbCV + ^H Z S 

49. SbCV + H 2 = SbOCI,, + 2H+ + 5C1- 

50. ISbOCl + H 2 S = Sb 2 S 3 + 2H 2 + 2H+ + 2C1~ 


L. Tin 

51. SnCl" 2 + Fe = SnCl^ 2 + Fe+ 2 + 2C1~ 

52. SnQ 4 2 + 2Hg' 2 + 4C1~ = SnCl' 2 + Hg 2 CI % 
SnCl 4 - 2 + Hg 2 C1 2 = SnC1 6 2 + 2H gz? 

M. Oxidizing Agents with H 2 S and H+ 

53. Cr 2 0-; + 8H + 3H 2 S = 2Cr+ 3 + 7H 2 O + 3S 

54. 2Fe+ 3 r + H 2 S = 2Fe+ 2 + S + 2H 1 

55. 2Mn0 4/ , u + 6H< + 5H 2 S = 2Mn^ 2 + 8H 2 O + 5S 

56. 2N0 3 + 2W + 3H 2 S = 2NO + 4H 2 O + 3S 

N. Organic Reagents with Group 2 Ions 

See Chapter 12. 


(A) If using a solution known to contain only the metals of group 2, 
start with 1 ml of sample and add 5 drops of H 2 O. 

If using a general unknown* adjust the volume of the group 1 ccntrate 
to 2 ml by adding H 2 O or boiling off as needed. 

Both of the above solutions are probably acidic already, but it is 
necessary to adjust the pH to 0.7-0.5 (0.2-0.3 M H f ), which is optimum 
for the precipitation of group 2 sulfides and the simultaneous prevention 
of precipitation of group 3 sulfides. If the solution is acidic as shown 
by litmus or other indicator, it is best to make the /;H about 7 with 6 M 
NH 4 OH, then to add dropwise, with good mixing, dilute HC1, checking 
after each drop the /?H result with methyl violet paper. This is done by 
touching a small drop of the mixture on a stirring rod to the indicator 
paper laid on a spot plate and comparing the color to that developed on 
the paper with a solution made from 1 ml of H 2 O plus 1 drop of 6 M 
HCl.f Do not put the paper itself in either solution. Solution volume 
at this point should be about 2.5 ml. 

* Certain ions such as NHJ and Ac~ slow the attainment of pH adjustment due to 
their buffering action. They may be removed by evaporation with HNO 3 followed by 
resolution with dilute HC1. These ions are assumed absent here, although pH attain- 
ment by use of methyl violet as described is valid in their presence. See list of reagents 
in appendix for preparation of this indicator paper. 

t Other methods for checking the /?H employ nitrazine paper or indicators like 
malachite green which are used directly in the solution. Adjustment of pH without 
special indicators may be approximated by making the solution just acid to litmus with 
NH 3 and HC1 as required, followed by addition of 1 drop of 6 M HC1 for every ml of 
solution present. 


(B) Add 1 ml of 8 % thioacetamide (or saturate the solution at room 
temperature with H 2 S gas),* and let the tube stand in a boiling water bath 
for about 7 minutes. Then add 1 ml of H 2 O and 12 more drops of thio- 
acetamide and continue to keep the tube hot for 3-5 minutes more. 
Group 2 sulfides should be completely precipitated by this time and 
coagulated in a fluffy mass. 

(C) Centrifuge the mixture, and save the centrate if it may contain ions 
of groups 3, 4, and 5 for further analysis. The residue is washed by 
stirring with 1 ml of H 2 O to which has been added 1 drop of 0.5 M 
HC1, and, after centrifuging, the wash liquid is combined with the previous 
centrate for groups 3-5 analysis. (If the sample is known to contain 
only group 2 metals, both these centrates are discarded.) The centrates, 
if saved, should be made slightly alkaline with NH 4 OH as indicated by 
a drop of phenolphthalein in the solution, and stored in a corked tube 
for the group 3 procedure.f 

(D) The residue from (C) is stined with a mixture of 2 ml of H 2 O, 
1 drop of thioacetamide, and 12 drops of 6 M KOH and heated for 
3 minutes in a water bath and stirred 2 or 3 times during this interval. 
(CAUTION! Do not allow this caustic solution to splatter by direct 
heating in an open flame. Wash off any spillage immediately with plenty 
of water. Neutralize with dilute HAc and more H 2 O.) Centrifuge and 
save both residue and centrate. Wash, the residue twice with 10 drops of 
H 2 O each time, and combine washings with original centrate. The 
centrate may contain any or all of the following 2B ions: HgS 2 , AsS~ 3 , 
SnS^ 2 , SnS-' 2 , SbS~ SnO^ 2 , Sb(OH)~ Sn(OH)^ 2 , Sn(OH)~ 2 , AsO,", etc., 
as well as OH~ S 2 , HS~, Ac~, S^ 2 , and NH 4 OH. Label this solution 
2B and save it for procedure (K). 

(E) The residue from (D) may contain any or all the 2A metals as 
sulfides: HgS, PbS, Bi 2 S 3 , CuS, CdS. To this washed residue, add I ml 
of H 2 O and 1 ml of 6 M HNO 3 and heat the mixture with stirring in a 
water bath for 3 minutes. Centrifuge. If the residue from this step is 
spongy and tends to float as a lump, it is sulfur and may be removed and 

* If oxidizing agents are present, H 2 S will reduce them and itself be oxidized to 
H 2 O t S. Four common oxidizing agents are: Cr 2 O 7 2 , MnO 4 , NO 3 and Fe+ 3 . These 
respectively give Cr' 3 , Mn+ 2 , NO and Fe 2 at this point, and do not interfere with the 
analysis except to require additional H 2 S to compensate for that oxidized. These 
metallic ions are all subsequently precipitated in group 3 and analyzed there. It is 
assumed these are not in the student sample in large quantity. 

t The [S~ 2 ] does not increase appreciably at room temperature in this solution and 
oxidation to SO^ 2 , which would precipitate group 4 metals prematurely, is not observed. 
The solution must not be made much more basic than/H 8.3 or Mg(OH) 2 wjll precipitate. 
This is undesirable prior to group 5 testing. 


discarded.* If a definite black precipitaie is left it is HgS, and a white 
precipitate is probably 2HgS Hg(NO 3 ) 2 . This is combined with HgS 
also to be found in the 2B centrate and is analyzed to confirm Hg in 
section (M) below. Finely divided sulfur may also appear as a white 
precipitate here. 

(F) The centrate from the HNO 3 solution may contain Pb+ 2 , Bi m , 
Cu+ 2 , and Cd+ 2 . Transfer it to a 20-ml beaker, add 6 drops of concen- 
trated H 2 SO 4 , and evaporate carefully under the hood with stirring and 
low, direct heat until copious white fumes of SO 3 are emittedf and only 
about 6 drops of liquid plus moist crystals are left. Allow the beaker to 
cool to room temperature, then dropwise, carefully add 1.5ml of H 2 O 
with stirring to avoid splattering.^ If a permanent, white, crystalline 
precipitate remains, it is probably lead sulfate and is tested in the next 

((7) Transfer the entire mixture to a test tube and rinse the beaker 
contents into that tube with 10 drops of waier. Centrifuge and pipet 
the supernatant liquid to another tube for further testing of 2A metals as 
directed in section (//). Wash the residue with 5-6 drops of H 2 O and 
reject the washings. Add 15 drops of NH 4 Ac, warm the mixture on a 
water bath and stir it. After several minutes the residue should have 
mostly dissolved. Centrifuge if any remains and test the centrate with a 
few drops of K 2 CrO 4 . A yellow precipitate of PbCrO 4 confirms Pb (as 
also obtained in group 1.) 

(H) The supernatant liquid from the first part of (G) may contain Bi m , 
Cu+ 2 , Cd+ 2 , and H 2 SO 4 . With stirring, add 15 M NH 4 OH dropwise 
until the solution is definitely basic, as evidenced by odor and litmus 
paper testing; then add 1 drop more. A deep blue color indicates the 
presence of Cu(NH 3 )+ 2 and a white, gelatinous precipitate is Bi(OH) 3 . 
Centrifuge. Save the residue for (/), the centrate for (/). 

(/) In another tube, mix 2 drops of SnCl 4 2 with 4 drops of 6 M NaOH 
and shake. An initial precipitate of Sn(OH) 2 dissolves to give the 
stannite ion, Sn(OH) 4 2 . Pour it over the residue saved from (//); a 
black precipitate of metallic bismuth shows this metal was in the original 

(J) The centrate from (H) contains copper if the solution is blue and 
no copper if the solution is not blue. If copper is not present, one may 
proceed directly to test for Cd by adding 6-8 drops of CH 3 CSNH 2 , 

* Some S~ 2 is oxidized to SO' 2 here but in a solution of this high ionic strength, 
PbSO 4 does not precipitate. 

t HNO 3 must be driven off here, since in its presence PbSO 4 is incompletely pre- 

% H 2 SO 4 is normally added to water; here the reverse is necessitated. 


6-8 drops of H 2 O, and heating 5-10 minutes in a water bath. A yellow 
sulfide, CdS, is sufficient evidence, since this is the only yellow sulfide in 
group 2 that is insoluble in 6 M KOH. If copper is present, it is necessary 
to remove or complex it in such a way that it does not interfere with tests 
for cadmium. As directed by the instructor use one of the methods 
(tests 16-5, 16-6, 16-7) given in the preliminaries, starting at the sign * 
in each case. 

This is the end of the 2A procedure. 

(K) To the alkaline solution and washings from (Z)), add a drop of 
methyl orange and enough 6 M HC1 to change the color to orange, 
pH ^ 4, then put in 3 drops of thioacetamide and heat in a water bath 
5-8 min with occasional agitation. Centrifuge, discard the centrate, 
and save the residue for (L). If the solution was only milky before 
centrifuging and contained no well defined residue, only very small amounts 
of 2B ions can be present.* If this is suspected, the solution is evaporated 
with 10 drops of dilute HNO 3 , the residue dissolved in dilute HC1, divided 
among several tubes, and tested according to the preliminary experiments 
for As, Sb, and Sn using the Gutzeit, Fleitmann, rhodamine-B, and flame 
tests which are sensitive enough to detect small amounts. 

(L) The residue from (K) may contain HgS, As 2 S 3 , Sb 2 S 3 , SnS 2 , and S. 
The color should give the student some clue to the metals present. Treat 
the solid with 1 ml of 6 M HC1, stir, and heat for 3-4 min; then centrifuge. 
The residue may contain black mercuric sulfide and/or yellow arsenous 
sulfide and sulfur. The centrate may contain SbClj" and SnCljr 2 . Segre- 
gate the clear centrate in another tube and test it according to section (0). 
The HgS-As 2 S 3 -S mixture is washed with 1 ml of H 2 O, the wash liquid 
rejected, then mixed with 10 drops of 6 M NH 4 OH, 5 drops of 3% H 2 O 2 
and heated with stirring for 4-5 min in a bath. After this interval, add 
1 ml of H 2 O, stir, and centrifuge. A black residue shows HgS; the 
clear centrate may contain AsO^ 3 . Save the residue for (A/), carry the 
centrate to (N). 

(M) The HgS from (L) is washed into the tube containing the mercury 
residue from (). Discard the wash liquid and to the solids add 8 drops 
of concentrated HC1 and 3 drops of concentrated HNO H , and heat and 
stir the tube contents in a water bath. After 3-4 min, transfer the solution 
to a 10-ml beaker and evaporate it slowly under the hood to a moist 
residue; do not evaporate it to dryness. Add 8 drops of H 2 O and centri- 
fuge if not clear. To the centrate add 2-3 drops of SnCl^" 2 solution. A 

* Some (finely divided) sulfur always shows up when an alkaline sulfide is acidified. 
Sometimes a sponge of sulfur floats to the top leaving a clear solution. If student 
samples are known, however, to contain 5-20 mg/ml of each ion, this evidence is 
indicative that no 2B metals are present. 


white to grey precipitate of mercurous chloride and mercury shows Hg+ 2 
was in the unknown sample. Alternately one may use the Reinsch 
test, 16-4. 

C/V) The AsOj 3 centrate from (L) is mixed with 6 drops of magnesia 
mixture.* A white crystalline precipitate of (NH 4 )Mg(AsO 4 ) indicates 
arsenic.f It is confirmed with a Gutzeit test or a Fleitmann test or these 
can be done directly on the unknown as spot tests. (Antimony gives the 
Gutzeit only, if also present.) 

(0) The centrate of soluble chloro complexes of antimony and tin 
from (L) is boiled gently for several minutes, without splattering, to hydro- 
lyze any CH 3 CSNH 2 remaining and to volatilize any H 2 S left. Dilute the 
solution to a total volume of about 2 ml, divide it between 2 tubes and 
test them as follows: 

Tube 1 : add a few iron filings, 2 drops of concentrated HC1, and heat. 
The iron will reduce the tin to Sn 11 and the antimony to Sb, the latter 
appearing as black flecks. To one portion of clear reduced solution, add 
1-3 drops of HgCl 2 and watch for the evidence of mercury reduction as 
noted in cation group 1 and in procedure (A/) above. With another 
portion perform the tin flame test as outlined in test 16-14. (The former 
method is better for estimating amounts as measured by the volume of 
mercury precipitate.) 

Tube 2: if no tin was found, dilute the solution to 3ml with H 2 O, 
add 4-5 drops of CH 3 CSNH 2 , and heat in a bath. An orange precipitate 
is Sb 2 S 3 and is sufficient evidence to say antimony is in the sample. If tin 
was found, run the following two tests which work in its presence: 

(a) Perform the rhodamine-B reaction as described in test 16-12 using 
a spot plate and a few drops of solution. Save the rest of the solution for 
part (b). 

(b) To the remaining solution add 10 drops of (NH 4 ) 2 C 2 O 4 and 4 drops 
of CH 3 CSNH 2 and heat in a water bath. The oxalate temporarily com- 
plexes Sn lv so that Sb 2 S 3 will precipitate alone.J In the absence of 
arsenic, antimony may be confirmed by the Gutzeit test directly on the 

This is the end of the group 2B procedure. 

* See list of reagents, appendix A 13. 

t Phosphates give the same test and are assumed not present in simple unknowns. 
Directions in Chapter 21 provide for the detection and elimination of PO^ 3 as well as 
giving other pointers on handling the general unknown. 

J If heating is prolonged, say beyond 5 minutes, some Sn ir is formed by the reducing 
action of H 2 S and may precipitate as brown SnS. 



1 . An acidic solution containing only salts of group 2 metals gives a black 
sulfide precipitate which is partly KOH soluble. The insoluble portion is 
solubilized completely with dilute HNO 3 , however, and when that solution is 
boiled down with H 2 SO 4 and then diluted, no precipitate is visible and the 
solution is colorless. Explain the significance of these observations both as to 
which metals might be, and which probably are not, present. Which have not 
been tested ? 

2. Show by calculation, how many drops (20 drops = 1 ml) of 8% thioaceta- 
mide solution are needed to precipitate 10 mg of Cu as CuS. 

3. If a solution is 10~ 2 M in CuCNHg)^ 2 and 10~ 3 M in uncomplexed NH 3 , 
show whether ~>r not CuS will form if to 20.0 ml of this mixture is added I drop 
of 0.5 M Na 2 S. Neglect volume change. 

4. Show by calculation whether or not PbS will precipitate from a solution 
saturated with PbSOj and having a /?H of 1.5, if treated with H 2 S gas until 

5. With the aid of the table of oxidation potentials, answer the following: 

(a) Is it possible to use Zn, Cd, Sn, and/or Pb to protect Fe from corrosion ? 
Of these, why is Zn used on galvanized fences and Sn on tin cans? 

(b) Suggest two reducing agents which would convert BiO+ to Bi and would 
be feasible for use in group 2 analysis. 

(c) From the two tin partials given with that element's description at the 
beginning of this chapter, explain why a piece of metallic tin is placed in the 
shelf bottle of SnCl 4 ~ 2 . 

(d) Comment on the tendency for cuprous ion to disproportionate in this way: 
2Cu f = Cu+ 2 4- Cu by evaluating the equilibrium constant for the reaction. 

(e) It has been suggested that if iron is used to treat an acidified mixture of 
Cu+ 2 and Cd +2 , only copper is reduced, leaving Cd +2 free in solution (with Fe+ 2 ) 
for testing. Is this separation feasible? If so, calculate the equilibrium 
constant for the reaction Cu f 2 4- Fe = Cu 4- Fe+ 2 and comment on its magnitude. 

6. (a) To an unknown suspected of containing only Bi 1IT and Cu +2 salts, one 
adds excess NH 4 OH. What information does this single reagent give? 

(b) A solution is suspected of containing only antimony salts. Why is H 2 S 
an excellent single reagent to use for a test ? What other tests would be indicated ? 

(c) A solution may contain Hg+ 2 and Cd+ 2 . What information does one 
elicit by adding some metallic magnesium and later, H 2 S ? 

7. Jackson P. Slipshod gives the following method for testing an unknown 
solution thought to contain Hg+ 2 , Pb+ 2 , and Cd+ 2 : 

"Add a piece of copper to plate out mercury. To that solution, preci- 
pitate lead as the sulfate, then add H 2 S to the centrate from the PbSO 4 
separation, and throw out yellow CdS." 

8. Could the pH for the initial group 2 precipitation conditions be correctly 


indicated by any of these: methyl orange, methyl red, malachite green. See 
chart of indicators, Appendix A 17. 

9. (Library) Find reference to the Marsh and Bettendorf tests. Describe 
them briefly and contrast them to a similar test in this chapter. 

10. Balance: 

(a) Bi(OH) 3 + Sn(OH)r 2 = Bi + Sn(OH)^ 2 

(b) CuS + H 3 0+ + NOg- = Cu+ 2 + S + H 2 + NO 

(c) As 2 S 3 + NOi" + H+ + H 2 = H 3 As0 4 + S + NO 

(d) H 2 S + Cr 2 7 ~ 2 + H+ = Cr+ 3 + H 2 O + S 

1 1 . Name as Werner compounds : 

Na 2 [Sn(OH) 6 ], (NH 4 ) 3 [SbS 3 ], [Cd(NH 3 ) 4 ]SO 4 , K 2 [Cu(CN) 3 ], H 2 [SnCl 4 ] 

12. (Library) Look up the composition and one use of each of the following 
materials that employ group 2 metals and their compounds: 

(a) constantan, (b) German silver, (c) Babbitt, (d) dental amalgam (e) Wood's 
metal, (/) Bordeaux mixture, (g) Paris green, (h) Fehlings' solutions, (/) tartar 
emetic, (/) bluestone. 

13. (a) Chronic arsenic poisoning in humans tends to concentrate the metal 
in hair roots. You are given a tuft of hair from an autopsy. How will you 
proceed to test for arsenic? 

(b) You are asked by a paint company to test a rival's yellow paint to see if the 
pigment is lead chromate or cadmium sulfide. What will you do? 

(c) How will you differentiate between a brass (Cu-Zn) and a bronze 
(Cu-Zn-Sn) by chemical means? Organize the proposal as a flow sheet. 

14. Using simple procedures such as a single solvent, observation, or reaction 
reagent, how would you differentiate between samples of these: 

(a) HgS and CuS, (b) SnCl^ 2 and SnClT 2 , (c) CdS and SnS 2 (d) AsCl 3 and 
CuCl 2 , (e) BiCl 3 and CdCl 2 , ( f) CdS and S, (#) air containing H 2 S and air con- 
taining SO 2 , (h) Pb and Cd metals, (/) Arsenic and antimony metals, ( / ) 
NH 4 C1 and NH 4 (OH), (using a metallic ion from group 2). 

15. In the group procedure, explain why these steps are taken: 

(a) The solution is boiled down with H 2 SO 4 prior to the separation of lead as 
PbSO 4 . (b) The solution containing Sb 111 and Sn IV is divided into several 
portions rather than separating one element from the other and proceeding with 
the centrate as one does for the other elements, (c) H 2 O 2 is added with NH 4 OH 
in treating a mixture of As 2 S 3 and HgS. (d) SnCl^ 2 is used instead of SnCljj" 2 
to test for Hg +2 . (e) KOH is added to the original group 2 residue. 

16. Which group 2 ion(s): (a) give black sulfides, (b) give yellow sulfides, (c) 
could be reduced in acid solution to their metallic state by using iron filings, 
(d) give a yellow sulfide with one oxidation state and a brown sulfide with 
another (e) appear in both groups 2A and 2B? 

17. What single reagent will dissolve one and not the other in each pair? 
Which is the soluble substance in each case ? (a) HgS and As 2 S 3 , (b) Cu(NH 3 ) 4 SO 4 
and CuS, (c) BiO(OH) and BiCl 3 , (d) Pb(OH) 2 and SnO 2 -H 2 O, (e) HgS and 
Bi 2 S 3 . 


18. Precipitation of As 2 S 3 from acid solutions using H 2 S gas frequently leads 
to a colloidal sulfide suspension, but this is not generally noted if CH 3 CSNH 2 is 
used as an H 2 S source. Explain the principle behind the advantage in using the 
organic precipitant. 

19. A solution is treated under precipitating conditions for group 2 analysis 
and a black precipitate is obtained. This is insoluble in KOH but soluble in 
dilute, hot HNO 3 giving a colorless solution. When the solution is boiled down 
with H 2 SO 4 and then diluted with H 2 O, only a clear solution is obtained. What 
group 2 ion(s) may be present? Explain. 

20. Explain these observations: 

(a) An acid solution of Sb lir is diluted with water and a white precipitate forms. 

(b) In the Cd-Cu separation using the cyanide method, a yellow color 
develops in the solution but no yellow precipitate. 

(c) A metallic sample suspected of being pure tin, gives a white residue when 
boiled in concentrated HNO 3 . 

(d) A mixture suspected of containing Cu +2 and Cd +2 is colorless when 
made basic with NH 4 OH. When Na 2 S 2 O 4 is added and the solution boiled, 
no reaction is apparent. 

O) A yellow group 2 sulfide is soluble in either dilute KOH or HC1. 

21. Heating changes potassium stannate from a soluble to an insoluble 
compound. On the basis of this alone, which formula for the compound do 
you like best and why: K 2 [Sn(OH) 6 ] or K 2 SnO 3 -3H 2 O? 

22. (d) Bismuth hydroxide is insoluble in 6 M KOH as well as in 6 M NH 4 OH. 
What can one deduce concerning the amphoteric character of this compound 
and tendency of Bi m to form complexes? 

(b) (Library) Describe the thiourea test Yor bismuth. 

23. Cu 2 [Fe(CN) 6 ] is soluble in NH 4 OH. Which furnishes fewer Cu+ 2 ions 
in solution, saturated cupric hexacyanoferrate(II) or tetrammine copper(ll) ion? 
Explain. PbSO 4 is soluble in NH 4 Ac, and if CrO^ 2 is added to that solution, 
PbCrO 4 precipitates. Which lead compound is most insoluble and why? If 
H 2 S is bubbled into a suspension of PbCrO 4 , the mixture begins turning black. 

24. A solution contains some group 2 ions. At pH 0.7, H 2 S gives a black 
precipitate that partly dissolves in KOH, giving residue A and centrate B. A is 
soluble in HNO 3 . Boiling this with H 2 SO 4 and diluting gives a white material, 
C. The centrate from removal of C gives a white precipitate with NH 4 OH, D, 
and a colorless centrate with which H 2 S gives a yellow solid, E. B gives a yellow 
precipitate with dilute HC1 which dissolves in excess HC1, and this, when 
caused to react with iron metal, yields a clear solution, F, which in turn reacts 
with HgCl 2 . Identify C, D, E, and F. 

25. In the discussion of group 1 in Chapter 15, two methods were used to 
abbreviate the procedure: a "flow sheet" and a "block outline." Prepare a 
block outline for group 2 by following the group 1 example and the group 2 flow 
sheet found in this chapter. 



1. J. T. Dobbins and E. S. Gilreath, /. Chem. Educ., 22, 119 (1945). (Non-H 2 S 

2. J. C. Herndon, /. Chem. Educ., 23, 183 (1946). (Groups 2 and 3) 

3. P. C. Gaines and R. Woodriff, J. Chem. Educ., 26, 166 (1949). (Dithionite for 
Cu+ 2 -Cd+ 2 ) 

4. M. V. Davis and F. H. Heath, J. Chem. Educ., 26, 277 (1949). (Cyanate for 
Cu+ 2 -Cd+ 2 ) 

5. H. E. Gunning, /. Chem. Educ., 32, 258 (1955). (Thioacetamide) 

6. G. F. Grillot and J. B. Kelley, Anal Chem., 17, 458 (1945). (Cu+ 2 -Cd+ 2 ) 

7. H. H. Barber and E. Grzeskowiak, Anal. Chem., 21, 192 (1949). (Thioacetamide) 

8. R. K. McAlpine, Anal. Chem., 25, 331 (1953). (NaCN on Cu(NH 3 )+ 2 ) 

9. H. Yagoda, Ind. Engr. Chem., Anal. Ed. 9, 79 (1937) (Confined spot tests) 

10. P. W. West, M. M. Vick, and A. L. LeRosen, Qualitative Analysis and Analytical 
Chemical Separations, Macmillan, New York (1953). 

11. J. Carnog, Semimicro Qualitative Analysis, Houghton-Mifflin, New York (1948). 



Five of the group 3 metals are transition elements, so-called because the 
filling of the 3d level starting with Sc and ending with Cu in the first long 
horizontal row results in a transition from a 2, 8, 8 electron foundation to 
2, 8, 18, while the 4s level holds at 1 or 2 electrons (see Chapter 3). The 
periodic relationships of the elements to be described are shown below. 



























FIG. 17-1. The periodic table in the vicinity of the group 3 metals. 


Aluminum is a strong, ductile, and malleable metal having good electric 
conductivity and excellent light reflectivity. The metal and its alloys are 



used in many applications where low density and good resistance to 
atmospheric corrosion are desired, such as aircraft skins and members, 
window frames, auto parts, cooking utensils, metal foils, and metallic 
paints. Alloys containing small percentages of Mn, Cu, Mg, and some- 
times Cr and Zn, have good tensile properties. See Tables 21-3 and A23. 
These wrought alloys are widely used in airplane manufacture; they can 
be age-hardened and are about twice as strong as pure aluminum. The 
latter's sp. gr. is 2.70, its m.p. 658 C, and b.p. is 2330 C. 

Aluminum is the third most abundant element, preceded only by oxygen 
and silicon, and these three elements are found combined in most igneous 
rocks, micas, feldspars, and clays. The most important mineral for metal 
recovery is the hydrated oxide, A1 2 O 3 XH 2 O, called bauxite. Iron, 
titanium, and silicon oxides are undesirable impurities in bauxite and are 
removed in the Bayer process by dissolving the aluminum oxide as aluminate 
ion, Al(OH)^f, in hot caustic soda, and later diluting the solution to reverse 
the reaction and precipitate hydrated aluminum oxide which is washed 
and calcined to make metal grade aluminum oxide. The purified oxide 
which represents about half the weight of crude ore is dissolved in cryolite, 
Na 3 AlF 6 , at high temperature and electrolyzed at about 1000 C in a 
Hall-Heroult or a Soederberg cell. Carbon anodes are used and the iron 
box of the cell is lined with carbon which serves as the cathode. Oxygen 
is liberated as the by-product, and molten aluminum is tapped off the 
bottom for casting into pigs of 99+ % Al. A large cell may produce a 
ton of Al and consume a half ton of anode daily. 

The only oxidation state of Al stable in water solutions is (III), though 
evidence of a lower state is obtained by anodic oxidation of the metal in 
various solvents. For example, after the current is shut off in the electro- 
lysis of Al in aqueous NaCl solution, H 2 is liberated at the anode, pre- 
sumably by Al 1 (formed in electrolysis) which reduces H+ to the gas and is 
thereby oxidized to Al 111 . The metal is strongly electropositive, though the 
adherent A1 2 O 3 coating which is present in normal application deactivates 
the metal greatly. Devoid of this, Al dissolves readily in either acids 
(except HNO 3 ) or bases: 

AI = Al+ 3 + 3<r E = 1.66 

Al + 30H- = A1(OH) 3 + 3e~ E^ = 2.31 

If the surface of the metal is amalgamated to keep it active, Al can be 
dissolved in water to give A1(OH) 3 (hydrated) or in alcohols to give 
A1(OR) 3 , aluminum alkoxides, which are strong bases. (See test 22-23 for 
an analogous borate ester.) Aluminum powder burns in air when heated 
forming a mixture of the oxide and nitride. It is capable of reducing oxides 


of less noble metals like iron, and use of this is made in thermite welding 
(Goldschmidt process) and thermite incendiaries. 

2A1 + Fe 2 3 = A1 2 3 + 2Fe + H r 

The heat of reaction H r easily renders the iron in the molten state. When 
heated with halogens or sulfur, aluminum forms the halides and sulfide 
A1X 3 and A1 2 S 3 ; the latter hydrolyzes in H 2 O to give H 2 S and A1(OH) 3 . 
Aluminum hydroxide (K SP = 5 X 10~ 33 ) is amphoteric, dissolving in 
acids to form salts when it functions as a base, and in strong bases to form 
aluminates when it functions as an acid : 

A1(OH) 3 + 3H+ + 3NO- = 3H 2 O + A1+ 3 + SNOg 
AI(OH) 3 + Na+ + OH- = Na+ + Al(OH)j- . 

Aluminum hydroxide is not soluble in NH 4 OH nor does Al 43 form an 
ammine complex, although ammoniates such as A1C1 3 -6NH 3 (corres- 
ponding to hydrates A1C1 3 -6H 2 O) are known. The nitrate, sulfate, and 
fluoride of Al are clearly electrovalent compounds as shown by their 
high melting points and conductance in aqueous solution, but the other 
halides of Al are largely covalent. Of these, aluminum chloride is indus- 
trially important as a catalyst in some organic reactions; it, for example, 
promotes the change from a straight-chain to branched-chain structures 
in alkanes, the latter having higher octane ratings. The aluminum halides 
are hydroscopic and decompose when heated to dehydrate them, giving 
A1 2 O 3 plus the halogen acid. The molecular weights in the vapor state 
correspond to a doubled molecule presumably held together through 
halogen bridge bonding: 


\ / ^ / 

Al Al 

/ \ / \ 

Complexes in solution range from A1X+ 2 to AIX^ 3 . K ins for AlFg 3 is 
1.44 x 10- 20 . 

Aluminum sulfate, A1 2 (SO 4 ) 3 -18H 2 O and the double sulfate, potassium 
aluminum sulfate (alum), K 2 SO 4 -A^SO^^HgO are two important 
commercial compounds of the metal and are prepared from reaction 
between clays and sulfuric acid. The compounds are both used because 
Al 111 hydrolyzes to give an acidic solution and also because (hydrated) 
A1(OH) 3 precipitates. 

A1+ 3 + 6H 2 = A1(H 2 0)+ 3 = A1(OH) 3 (H 2 0) 3 + 3H+ 
Thus one use of aluminum sulfate is in baking powders where its acidity 


causes NaHCO 3 to release CO 2 , which raises dough; another use is in 
settling suspended matter in municipal water treatment by adsorption and 
mechanical entrapment with the flocculant aluminum hydroxide 

Qualitatively, Al +3 is identified by the characteristic gelatinous appear- 
ance of its hydroxide, and, in lower concentration, by formation of a red 
lake with the organic reagent, aluminon (test 17-6 this chapter, and 
Example 16, Chapter 12). Quantitative determination is similar; large 
quantities are precipitated with NH 4 OH and ignited and weighed as the 
oxide, or precipitated with oxine (special experiment 7, part 6). Small 
quantities are measured colorimetrically using aluminon. 


This metal is hard and brittle and has a high luster. It has two crystal 
forms, a hexagonal form stable below 800 C, and a cubic form stable at 
higher temperatures. Of the elements in periodic group VIB, chromium 
is most abundant and easiest to win from the ore. The chief mineral is 
chromite, Cr 2 O 3 -FeO which is reduced by heating with coke to give an 
iron-chromium alloy that is added as such to iron melts to give chromium 
steels. These steels contain 14-30% Cr. Two common ones are: 
18 % Cr, 8 % Ni, remainder iron (called 18-8 stainless) for applications like 
the construction of food handling machinery; and 15% Cr, 60% Ni, 
remainder iron (called nichrome) and used in heating elements. Such 
alloys are tough and difficult to fabricate but have good corrosion resistance. 
Considerable Cr 2 O 3 is used to make furnace linings and green paints. 
Cr metal is also used as a protective and decorative coating on other 
metals, as in auto fixtures. The sp. gr. of Cr is 7.1, the m.p. is 1540 C, 
and the b.p. is 2475 C. 

The oxidation states of chromium are (II), (III), (IV)?, (V), and (VI). 
Chromium may also be said to have a (0) oxidation state in chromium 
carbonyl, Cr(CO) 6 , a covalent compound of considerable stability. Dona- 
tion by CO of a total of 12 electrons gives an electron configuration about 
Cr like Kr, the next rare gas. Mo and W also form hexacarbonyls. The 
metal dissolves slowly in acid to give chromous ion, Cr 11 , which is readily 
oxidized to the chromic state, Cr 111 . The chromic ion has a strong ten- 
dency toward 6-coordination with ions like halide and cyanide and is 
present in water solution as characteristically violet-colored Cr(H 2 O)J 3 . 
Three partials are 

Cr = Cr+ 2 + 2e~ E = 0.91 

Cr = Cr+ 3 + 3e~ E = 0.74 
Cr+ 2 = Cr+ 3 + e~ E = 0.41 


Heated Cr metal reacts with O 2 to give Cr 2 O 3 , with N 2 to yield CrN, 
with halogens to give CrX 3 (except in the case of I 2 where the latter's 
lack of oxidizing strength gives only CrI 2 ), and with sulfur to give CrS 
for the same reason. Chromous compounds somewhat resemble corres- 
ponding ferrous compounds but are not important. Adding base to 
solutions of chromic salts precipitates hydrated chromic oxide, written 
Cr(OH) 3 (K$p = 7 x 10~ 31 ), although no definite formula may be assigned. 
This hydroxide is amphoteric and dissolves in either strong acids or bases, 
the latter giving chromites that have variable composition depending upon 
base strength : 

Cr(OH) 3 + 3H+ = Cr' 3 + 3H 2 O 

Cr(OH) 3 + 3OH~ = Cr(OH)- 3 

Concentrated NH 4 OH solutions give Cr(NH 3 )+ 3 . In addition to the 
oxide and hydrous oxide (hydroxide), the phosphate of Cr TTI is also only 
slightly soluble. 

Cr VI compounds are industrially important. The chromates, CrO 4 2 , 
and dichr ornate s, Cr 2 O^ 2 , are the forms in which hexavalent chromium is 
known. Cr YI is a powerful oxidizing agent in acid solution. 

2Cr 13 + 7H 2 O = Cr 2 O7 2 + 14H+ + 6e~ E = - 1.36 

Advantage is taken of this in analytical chemistry in titration of Fe+ 2 
to Fe+ 3 , and in commerce for the tanning of hides. Na 2 CrO 4 - 10H 2 O 
is made by fusing chromite ore with soda ash and extracting the mass with 
water. Na 2 Cr 2 O 7 2H 2 O is prepared from that solution by acidification 
with H 2 SO 4 . 

The chromic-chromate-dichromate-perchromate transformations are 
illustrated in the group 3 analytical procedure. In basic solution, green 
Cr(OH) 3 is converted to Cr(OH) 4 and oxidized to yellow CrO 4 2 by 
H 2 O 2 , since at high /?H, the potentials allow hydrogen peroxide to convert 
Cr 111 to Cr VT , a reaction that goes the other way in dilute acid solution. 

2Cr(OH) 4 + 20H- + 3H 2 O 2 = 2CrO 4 + 8H 2 O 

Acidification of this mixture with HNO 3 , followed by addition of H 2 O 2 
and butanol, C 4 H 9 OH, gives orange Cr 2 O^ 2 , then the unstable perchromic 
acid which dissolves in and colors the alcohol layer blue: 

2CrO 4 2 + 2H+ = Cr 2 Of a + H 2 O 
H 2 Cr 2 7 + 4H 2 2 = 2CrO 5 + 5H 2 O 

The color changes are used in qualitative testing to detect chromium. 

A covalently bonded Cr vl compound is chromyl chloride, CrO 2 Cl 2 , a 
liquid which is soluble in CC1 4 , CS 2 , etc., and used in certain organic 


oxidations. It is prepared by heating a dichromate, alkali metal chloride, 
and sulfuric acid together. Chloride is identified via this reaction due to 
the distinctive red color of CrO 2 Cl 2 . Chromyl chloride boils at 116C 
without decomposition but is easily hydrolyzed to H 2 O + C1 2 + CrCl 3 . 

Quantitatively, large amounts of chromium are put into solution by 
sodium carbonate sodium peroxide fusion, followed by acidification. 
Addition of KI to the Cr 2 C>7 2 yields an equivalent amount of I 2 , which is 
titrated with standard Na 2 S 2 O 3 . Small amounts (< lOppm) of hexa- 
valent Cr are run colorimetrically with s-diphenylcarbazide. 


Iron is the most important of the structural metals, and, fortunately for 
mankind, it is the fourth most abundant element in the earth's crust and is 
concentrated in several huge deposits. The sp. gr. of Fe is 7.86, its m.p. 
is 1539C, and b.p. is 2800 C. The human body is about 0.005% by 
weight iron, and igneous rocks are about 5.0% by weight iron. Gradual 
weathering of rocks (primarily calcium, magnesium, and iron silicates) 
has resulted in deposition of oxides of iron from which the metal is obtained 
by coke reduction, as explained in Chapter 13. These minerals include 
hematite, Fe 2 O 3 , magnetite, Fe 3 O 4 , and limonite, 2Fe 2 O 3 3H 2 O. Iron 
shows three distinct solid forms plus another which differs only in 
magnetism from one of the other three. The a-, /?-, and <5- forms are all 
body-centered cubical lattices, whereas y-iron is face centered; a and /? 
are similar except that a is magnetic and none of the others are. Little 
is known about (5-iron since it is stable only in a narrow range of high 
temperatures. The relationship is: 

770 C 928 C 1411 C 1530C 3235 C 

a^/J^y^<5^ Liquid ^ gas 

Uses for iron and its alloys are too numerous and obvious to mention, 
but it may be repeated here that the most important alloys are those with 
carbon (the steels) because the normally soft iron is greatly hardened by 
carbon, and steels are capable of wide variation of properties by heat 

The oxidation states of iron are (II), (III), (IV), and (VI); they are 
respectively illustrated by Fe+ 2 ferrous, Fe+ 3 ferric, FeOj 1 perferrite, and 
FeOj 2 ferrate. Of these the perferrites have been prepared only by high- 
temperature reactions of melts like Fe(NO 3 ) 3 + Sr(NO 3 ) 2 , whereas the 
ferrates, which are also strong oxidants, are made in basic solutions by 
anodic oxidation of iron or by chlorination of a suspension of hydrous 
ferric oxide. In addition to these four oxidation states, a (0) state is also 
observed in a few compounds like iron pentacarbonyl, Fe(CO) 5 , a liquid 


boiling at 102 C and prepared by the action of carbon monoxide on 
powdered iron. The compound is poisonous and its formation accounts 
for the disintegration of metal around automobile exhausts. For pur- 
poses here, only the common (II) and (III) states will be considered. 
Related couples in acid solution are: 

Fe = Fe+ 2 + 2e~ E = 0.44 

Fe+ 2 = Fe+ 3 + e~ E = - 0.77 

Iron metal is therefore attacked by acids and is capable of reducing Fe 43 
to Fe+ 2 . Strong oxidizing agents like Cr 2 Of 2 and HNO 3 passivate iron 
metal by giving it an oxide surface. Iron also dissolves in basic solution 
where ferrous hydroxide is readily converted to ferric which is insoluble 
and helps displace the equilibrium: 

Fe + 2OH- = Fe(OH) 2 + 2e~ = 0.88 

Fe(OH) 2 + OH- = Fe(OH) 3 + er = 0.56 

Ferrous hydroxide is noted to be a much better reducing agent than Fe+ 2 . 
Ferrous compounds are usually light green, since that is the color of Fe^ 2 . 
Ferrous sulfate heptahydrate (copperas), FeSO 4 7H 2 O, is the most common 
ferrous salt. It is made by the oxidation of the natural iron sulfide, 
pyrite or fool's gold, FeS 2 . Ferrous sulfate is used to make "iron blues," 
which are pigments produced by reaction of Fe+ 2 with ferrocyanide and 
followed by oxidation to Prussian blue with dichromate (see p. 403, Chapter 
22). These pigments are used to make inks (see special experiment 7) 
and other chemicals. Ferrous sulfide, FeS, is preparable by direct com- 
bination of the heated elements; it yields H 2 S when acidified. Ferrous 
ammonium sulfate (Mohr's salt), FeSO 4 (NH 4 ) 2 SO 4 6H 2 O, is used as a 
standard in quantitative analysis. 

Ferric compounds are normally obtained when ferrous compounds are 
oxidized and vice versa. Ferric oxide (hydrated), written zsferric hydroxide, 
Fe(OH) 3 , forms readily from Fe(OH) 2 under the influence of air oxidation. 
Dissolving the red-brown Fe(OH) 3 in acids yields ferric salts. An exception 
is HI: its reducing action gives FeI 2 as the final product. Fe 111 forms 
many soluble complexes: FeX^ 3 (with halogens), Fe(C 2 O 4 ) 3 3 , Fe ( CN )e 3 
Fe(SCN)g 3 , etc., some of which are illustrated under the sections on 
anions. Depending upon concentrations, intermediate amounts of 
coordination are possible, Fe(SCN)+ 2 , Fe(SCN)J, etc. Ferric oxide, 
Fe 2 O 3 , may be prepared in different shades of red and is used as a pigment; 
ammonium hexacyanoferrate(HI), (NH 4 ) 3 [Fe(CN) 6 ], is the source of blue 
color in blue print paper, and ferric chloride, FeCl 3 , is used as an acid 
etchant for metals. 



Reactions (in order of decreasing [Fe +2 ]) 


Formula Weight 

Fe(OH) 2 = Fe+ 2 + 2OH~ 

1.8 x 10- 15 


FeCO, = Fe+ 2 + CO^ 2 

2.11 x 10- 11 


FeS (a) = Fe+ 2 + S~ 2 

4 x 10~ 17 


FeSe = Fe+ 2 + Se 2 

io- 26 


FeCCN)^ 4 = Fe+ 2 + 6CN~ 

~ io- 35 



Reactions (in order of decreasing [Fe+ 3 ]) 


Formula Weight 

Fe(SCN)*- 2 = Fe^+ SCN~ 

1.04 x IO- 3 


FePO 4 = Fc^ 3 + POr 3 

1.5 x IO- 18 


Fe(OH)j - Fe+ 3 + 3OH~ 

6 x IO" 38 


FeF^ 2 = Fe+ 3 + 5F~ 

5 x lO" 16 


Fe S 3 = 2Fe+ 3 + 3S~ 2 

io- 88 


Fe(CN)^ 3 = Fe+ 3 + 6CN~ 

io- 42 


Ferrous and ferric iron can be differentiated simply by their color in 
water solution or the color produced with ions like SCN~, Fe(CN);r 3 , 
and Fe(CN)^" 4 as illustrated in test 17-1. Ferric ferrocyanide is called 
Prussian blue, ferrous ferricyanide is Turnbull's blue, ferric ferricyanide is 
Berlin green (though it is brownish colored), and ferrous ferrocyanide has 
no common name and is white, or if air oxidized, light blue. 

There are many sensitive organic reagents that are used in colorimetry 
for both qualitative and quantitative analysis', two good ones are 
1,10-phenantholine and a,a'-dipyridyl. The usual macro quantitative 
method for iron is one in which Fc f3 is reduced to Fe 42 with either Zn 
metal or Sn 11 and then reoxidized by titration in acid solution with 
MnOj, Cr 2 O 7 2 , or Ce TV . 


This is a silvery metal with a slight bluish color that is harder and 
stronger than either Fe or Ni and is magnetic below 1150 C. Its sp. gr. 
is 8.9, the m.p. is 1493 C, and b.p. is 3520 C. The principal minerals are 
cobaltite CoAsS, smaltite CoAs 2 , and cobalt bloom, 3CoO-As 2 O 5 -8H 2 O. 
These must go through several refining processes to recover cobalt metal 


from the metallic impurities which include Ag, Fe, Ni, and Cu. A mixture 
of NiO + CoO is obtained after several roasting steps, and this is handled 
by the Mond process in which reduction with CO + H 2 at 300 C gives 
the metals. Nickel is then removed from the powdered metallic mixture 
by causing it to react with CO and form nickel carbonyl, Ni(CO) 4 , which is 
volatile. The Ni(CO) 4 is pumped through a tower packed with nickel 
pellets where it decomposes reversibly at 180 C. Cobalt does not react 
with carbon monoxide. 

Cobalt metal is used in some important applications by itself and as an 
alloying metal. When bombarded in atomic piles it acquires and retains 
radioactivity better than any other common metal and is used as a 
radiation substitute for radium in cancer therapy. It may also be used in 
bomb cases containing nuclear explosives, since the explosion would 
produce radio cobalt vapor that might remain dangerous for years and 
would increase the casualty range with its radioactive "fallout." Cobalt 
is added to Cr Ni Mo C steels up to 12% by weight to give a product 
useful in cutting hard materials such as heat-treated steels. Cobalt 
powder is used as a binding agent in making sintered tungsten and 
titanium carbides, such as carbaloy, which are used in drawing and 
extrusion dies. The stellite alloys containing about 1.25% C, 65% Co, 
28% Cr, and 4% W are used in cutting tools because of their wear and 
oxidation resistance, ability to keep the edge at high temperatures, and 
low friction coefficient with respect to most metals. 

The oxidation states of Co are tl), (II) cobaltous, (111) cobaltic, and 
(IV). Of these the only simple ion stable in water solution is Co 12 
(hydrated). Co 1 and Co IV are rare, and Co 13 is a powerful enough 
oxidizing agent to liberate O 2 from H 2 O and is only capable of existence in 
aqueous solution in the form of complexes such as Co(NH 3 )J 3 . Some 
potential values of interest are: 

Co = Co+ 2 + 1e~ E = 0.28 

Co+ 2 = Co+ 3 + e~ E = - 1.84 

Co(NH 3 )+ 2 = Co(NH 3 )+ 3 + <r = - 0.1 

Co(H 2 O) 2 is pink. When a little strong base is added, a bluish 
Co(OH) 2 precipitates; more base changes this to a pinkish Co(OH) 2 and 
air gradually oxidizes it to black Co(OH) 3 . Cobaltous ion gives black 
CoS with alkaline sulfides. As freshly precipitated, it is slowly soluble 
in dilute HCl and is called the a-form, but after standing it changes struc- 
ture to the /S-form which is considerably less soluble. Cobaltous chlorides 
in which the coordination number of cobalt is 6, are pink, whereas those 
with 4-coordination are blue. Both Co 11 and Co m form many Werner 



compounds readily coordinating NH 3 , H 2 O, CN~, X~, NOj, etc. See 
Chapter 4 for structures. The coordination number is usually 6 and the 
geometry is octahedral. 

Cobalt compounds have limited use; for example, CoCl 2 is used with 
dessicants like silica gel as an indicator, since, when it changes from its 
blue color to the pink of [Co(H 2 O) 6 ]Cl 2 , it indicates that the dessiccant's 
absorptive capacity is low. CoO is used in ceramic glazes to give a blue 
color, as one might guess from the bead tests described in experiment 5, 
Chapter 23. Cobalt napthenates are used as paint and varnish driers. 

The blue borax bead is good qualitative evidence of Co, as is the forma- 
tion of yellow K 3 Co(NO 2 ) 6 , test 17-9. 


Reactions (in decreasing order of [Co 4 2 ]) 


Formula Weight 

Co(NH 3 ) 6 ^ 2 = Co+ 2 + 6NH 3 

1.25 x 10~ 5 


Co(OH) 2 = Co+ 2 + 2OH- 

2.5 x 10~ 16 


CoC0 3 = Co+ 2 + COyT 2 

8 x 10~ 13 


CoS (a) = Co^ 2 + S- 2 

5 x 10~ 22 


CoS (/3) = Co+ 2 + S- 2 

1.9 x 10- 27 


Quantitatively, Co may be determined colorimetrically with SCN~- 
acetone and extraction of the color, or larger amounts can be determined 
electrolytically after removing any iron as a phosphate precipitate and 
separating Co from any Ni by precipitation (and later resolution) using 
a-nitroso-^-naphthol. The subsequent electrolysis is done from NH 4 OH 


Nickel is a silvery metal which is malleable and ductile and yet stronger 
than iron. It does not oxidize as readily as iron and is magnetic below 
345 C. The sp. gr. is 8.9, the m.p. is 1452 C, and the b.p. is 2800 C. It 
is used in many corrosion-resistant alloys as monel, hastelloy, stainless 
steel, and in the important magnetic alnico alloys. See appendix A23. 
Nickel is also used as a plating metal and as a catalyst for hydrogenations, 
such as the conversion of cottonseed oil to solid cooking fats. 

The main minerals are pentalandite (2FeS NiS), nickelite, and millerite. 
The latter two are nominally NiS. Nickel is separable by the Mond 
process described under Cobalt or once isolated as NiS it is reducible by 
heating with coke. Purification is frequently by electrolysis. 


The chemistry of Ni is similar to that of Fe and Co though nickel is 
less active. 

Ni = Ni+ 2 + 2e- = 0.25 

The nickel oxidation states are (0), (I), (II), (III), and (IV). These are 
illustrated respectively by nickel carbonyl, Ni(CO) 4 , potassium tricyano- 
nickelate(l\ K 2 [Ni(CN) 3 ], nickelous sulfate hexahydrate, NiSO 4 -6H 2 O, 
nickel trioxide, Ni 2 O 3 , and nickelic oxide, NiO 2 . Not much is known 
about any of these states except the dipositive one, since the others are 
produced only under special conditions. NiO 2 is used in the Edison 
battery, NiO 2 + Fe + 2H 2 O = Ni(OH) 2 + Fe(OH) 2 reaction is in KOH 
solution. Most other compounds of the unfamiliar states are of but 
theoretical interest and for discussion here, the chemistry of nickel is that 
of Ni f2 and its compounds. Among these the oxide NiO, hydroxide 
Ni(OH) 2 , sulfide NiS, carbonate NiCO 3 , chromate NiCrO 4 , oxalate 
NiC 2 O 4 , and phosphate Ni 3 (PO 4 ) 2 are not appreciably water soluble 
although they will dissolve in mineral acids. NiS, like CoS, will not 
precipitate in acidic solution and once formed, is slow to dissolve in dilute 
HC1; a separation from other group 3 precipitates is sometimes made on 
that basis. NiS has three forms each possessing a different solubility 
product constant (Table 17-4). The a-form predominates in alkaline 
sulfide precipitations but changes to the less soluble /ft- or y-form upon 
standing or if acid is added. 

Hydrated Ni+ 2 salts are green and dehydrated ones are yellow. Nickelous 
ion, like the cobaltous ion, forms many complexes (though they are not 
as stable), such as Ni(CN) 4 2 , Ni(H 2 O)+ 2 , and Ni(NH 3 )+ 2 . In these, 
Ni f2 has the argon shell plus eight electrons in the 3d group. Coordina- 
tion of 4 groups, each contributing two electrons, establishes a 3d, a 4s> 
and two 4p orbitals resulting in a square planar spatial arrangement 
characteristic of dsp 2 hybridization. Ni+ 2 also forms complexes with 
6-coordinated groups, as Ni(NH 3 ) 2 , in which the 3d, 4s, and 4p quantum 
levels are filled by electron donation of six pairs of electrons from 
coordinated groups. 

Nickel and cobalt are generally isolated from other group 3 ions, then 
tested in the presence of each other, so tests for Ni+ 2 which are not inter- 
ferred with by Co+ 2 and vice versa are important analytically. Thus 
SCN~-acetone, a-nitroso-/?-naphthol, and KNO 2 HAc react with Co+ 2 
but not with Ni +2 . Dimethylglyoxime-NH 3 precipitates Ni+ 2 but not Co 42 . 

The characteristic red nickel-dimethylglyoxime chelate is used quantita- 
tively simply by filtering it off, drying at HOC, and weighing. Nickel 
is assayed volumetrically using standard KCN, which forms Ni(CN) 4 2 
in a solution made slightly basic with NH 4 OH ; Agl is used as the indicator 



Reactions (in order of decreasing [Ni+ 2 ]) 


Formula Weight 

NiCO, = Ni+ 2 + C0 3 2 

1.36 x 10- 7 


Ni(OH) 2 = Ni+ 2 + 20H- 

1.6 X 10- 16 


Ni(NH 3 )^ 2 = Ni+ 2 + 4NH 3 

1 x 10- 8 


Ni(NH 3 )^ 2 = Ni+ 2 + 6NH 3 

1.8 x 10- 9 


NiS (a ) = Ni+ 2 + S~ 2 

1 x 10- 22 


NiStf) = Ni+ 2 + S~ 2 

1 x 10~ 26 


NiS( y ) = Ni f2 + S- 2 

2 x 10- 28 


Ni(CN)i- 2 = Ni+ 2 + 4CN~ 

io- 22 


since when all the Ni+ 2 is complexed as the cyanide, the turbidity due to 
Agl disappears with the next drop of CN~. 

Agl + 2CN- = Ag(CN) 2 + I- 

Nickel may also be determined by electrolysis in NH 4 OH solutions 
previously freed of copper and cobalt. 


This metal has a reddish tint and is moderately soft if pure, but as 
normally obtained by carbon reduction of the dioxide it is hard due to the 
presence of Mn 3 C in solid solution. Electrolytic reduction gives pure 
metal. The sp. gr. is 7.2, the m.p. is 1244 C, and the b.p. is 2087 C. Man- 
ganese is used in at least three important general alloy types: (1) medium 
manganese structural steels containing 0.75-1.75% Mn, which are as 
strong as plain carbon steels yet contain less carbon and hence are more 
ductile (2) manganese bronzes, which are Cu-Zn brasses to which are 
added 0.1-5.0% Mn, 0.8-4.0% Fe, and 0.1-1.5% Sn to increase hardness 
and strength and decrease dezincification and grain growth and (3) Hadfield 
steels, containing up to 14% Mn, which have unusual wear resistance due 
to a work hardening of the surface yet exhibit retention of toughness in 
the rest of the structure. Hadfield steels find application in rail switches, 
earth moving equipment, conveyors, crushing machinery, etc., where wear 
is apt to be excessive. 

The important mineral is pyrolusite, MnO 2 , which is usually mixed with 
iron ores for common reduction to make ferromanganese, about 75% 
Mn and 25 % Fe, and Spiegeleisen, roughly the reverse composition. Both 
alloys are added to iron melts in the preparation of steels. Low-grade 
ores containing CaO, Fe 2 O 3 , SiO 2 , A1 2 O 3 , and MnO 2 are concentrated by 
the recently announced Nossen process in which the powdered ore is 


reduced at about 800 C to give MnO. The mass is extracted with HNO 3 , 
solubilizing Mn and Co and leaving the rest, from which iron can be 
obtained magnetically. The solution is concentrated and HNO 3 , 
Ca(NO 3 ) 2 (for fertilizer), and MnO 2 are recovered by heating the concen- 
trate. About 0.8 million tons of Mn metal are used yearly in the U.S. for 
steel making, and another 0.1 million tons of MnO 2 are consumed in the 
chemical and dry cell battery industries. Pyrolusite, interestingly enough, 
is found on the ocean floor as well as in more conventional deposits. 
These so-called "manganese nodules" have not yet been exploited. 

The oxidation states of Mn are (I) as K 5 [Mn(CN) 6 ], potassium hexa- 
cyanomanganate(\)\ (II) as MnSO 4 , manganous sulfate\ (III) as Mn(OH) 3 , 
manganic hydroxide \ (IV) as MnO 2 , manganese dioxide \ (V) as Na 3 MnO 4 , 
sodium manganate(V)\ (VI) as CaMnO 4 , calcium manganate, and (VII) as 
KMnO 4 , potassium permanganate. Of these, Mn 11 and Mn m have basic 
characteristics, Mn lv is amphoteric, and the three higher states are acidic. 
A//* 111 and A//? v compounds are noi stable toward disproportionation. 
Mn VT is stable only in basic solution and Mn 1 is known only in complex 
cyanides. Mn 11 is a reducing agent and Mn VTI is an oxdizing agent. 

Manganous ion, Mn f2 is the most familiar manganese ion in solution 
and most of its compounds are light pink. Bases precipitate Mn(OH) 2 
which does not dissolve in excess base but is soluble if the concentration 
of NH+ is high. Air oxidizes the hydroxide to a brown hydrous oxide of 
Mn u \ which is usually represented as Mn(OH) 3 . Heating any of these 
hydroxides in air gives Mn 3 O 4 . Manganese metal dissolves in acids to 
yield salts plus hydrogen, and is slowly soluble in bases giving the hydroxide 
and hydrogen: 

Mn = Mn+ 2 + 2e~ E = 1.18 

Mn + 2OH- = Mn(OH) 2 + 2e~ = 1.55 

Pink MnS is precipitated from Mn 4 2 solutions by alkaline sulfides and it 
is soluble in dilute acids. Other slightly soluble Mn+ 2 compounds are 
the carbonate, cyanide (soluble in excess due to complex formation), 
oxalate, and phosphate. None of these is of commercial interest. 

MnO 2 is the only important Mn lv compound. It is used in dry cell 
batteries (Zn = Zn+ 2 + 2e~ and 2MnO 2 + 4H 2 O + 2e~ = 2Mn(OH) 3 
+ 2OH~) and as a drier in paints. MnO 2 dissolves slowly in mineral 
acids to give Mn lv salts, but these are not stable and change to Mn 11 
with oxidation of the anion. MnO 2 fused with a base like KOH or 
Na 2 CO 3 to which has been added a strong oxidizing agent like KC1O 3 
gives green manganates as K 2 MnO 4 . Acidification brings about 
disproportionation : 

3MnO 4 2 + 4H+ = 2MnO 4 + MnO 2 + 2H 2 O 



In basic solution, in the presence of strong oxidizing agents, manganates 
are oxidized to permanganates. KMnO 4 , the most common of these, is 
prepared by fusing MnO 2 with KOH and oxidizing the manganate in 
KOH solution by bubbling in chlorine. Related half cell data are: 

Mn+ 2 
MnO 2 

The couple 
Mn+ 2 

2H 2 O = MnO 2 + 4H+ + 2e~ E = - 1.23 

4OH- = MnO 4 2 + 2H 2 O + 2e~ E = - 0.60 

MnOj 2 = Mn0 4 + e~ E = - 0.56 

2H 2 = Mn0 4 + 8H+ + 5e~ E = - 1.51 

s a familiar one in redox titrimetry. Oxidations using MnO 4 are generally 
run in sulfuric acid solution. Typical reducing agents quantitatively 
determined with standardized KMnO 4 are Fe+ 2 , H 2 O 2 , H 2 C 2 O 4 , HNO 2 , 
Sn n ,.Mo IIT , U 1V , VO+ 2 , As 111 , and Ti 111 . In these Mn+ 2 and the other 
products are not highly colored, so MnOj acts as its own indicator, the 
first drop in excess coloring the solution. If oxidations are carried out in 
neutral or basic solution, brown MnO 2 precipitates and the end point is 
not clear. With neutral solutions in the presence of Ba+ 2 , reaction goes 
only to Mn vl as BaMnO 4 precipitates, a reaction used to determine some 
organic reductants like alcohols, aldehydes, and cyanides which otherwise 
present difficulty, due to side reactions, if run in acid solutions. 


Reactions (in order of decreasing [Mn+ 2 ]) 


Formula Weight 

Mn(OH) 2 = Mn+ 2 + 2OH~ 

1 x 10- 19 


Mn 3 (PO 4 ) 2 = 3Mn+ 2 + IPO^ 3 

1Q -22 


MnCO 3 = Mn+ 2 + COg 2 

8.8 X 10~ n 


MnS = Mn+ 2 + S~ 2 

8 x 10- 14 


MnC 2 O 4 = Mn+ 2 + C 2 O 4 2 

1.1 x 10- 15 


Manganese is found qualitatively by oxidation, as previously mentioned, 
to MnO 4 whose purple color is distinctive, or to MnOj 2 whose green 
color is easily recognized in fusion mixtures. 

Quantitatively, manganese is oxidized to MnO 4 and small amounts are 
found colorimetrically. Larger amounts are handled by titrating with a 
standardized solution of reducing agent such as sodium arsenite. 



Zinc is malleable and ductile in the range 100-150 C but somewhat 
brittle otherwise. Its sp. gr. is 7.14, m.p. is 419.4 C, and b.p. is 907 C. 
The metal is used as a protective coating on iron fences, sheeting, and 
pipes. It is deposited by dipping in molten zinc (galvanizing), by 
baking on a layer of zinc powder (sherardizing), or by electrolysis in 
alkaline baths. Zinc is also used in several important alloys; brass 
is the one produced in greatest tonnages. Zinc dust is also used in 
certain metallurgical reductions to displace more noble metals from 

Of the six oxide, sulfide, and carbonate ores in demand today as zinc 
sources, sphalerite, sometimes called black jack, ZnS, and zincite, ZnO, 
are most widely distributed. The ores are concentrated and roasted to 
ZnO. The metal is then obtained by two general processes: (1) carbon 
reduction and distillation of zinc by either the older batch, Belgian 
horizontal-retort method, or the continuous vertical-retort method, or 
(2) electrolytically at 35 C from a dilute H 2 SO 4 -ZnSO 4 solution, using an 
Al cathode and Pb anode. The zinc obtained in the latter method is 
remelted and cast into slabs for further processing; the more pure zinc 
from the former method may be redistilled to greater than 99.99% 
purity. The usual impurities are cadmium (which zinc strongly resembles), 
lead, and iron. 

Experiments have indicated that zinc has oxidation states of (0), (I), 
(II), and (III), but almost all of zinc chemistry is that of dipositive zinc as 
the zinc ion, Zn +2 , and as the zincate ion. The latter varies in composition 
with base strength, is readily hydrolyzed by dilution, and written variously 
as ZnO 2 2 , HZnO 2 , Zn(OH)g, and ZntOH)^ 2 . Zinc metal is a good 
reducing agent and dissolves in acid or base: 

Zn = Zn+ 2 + 2e~ E = 0.76 

Zn + 2OH- = Zn(OH) 2 + 2e~ = 1.25 

The powdered metal burns readily in air to the oxide which is basic, 
Zn(OH) 2 is a crystalline material unlike the "hydroxides" of Al, Fe, Ti. 
Cr, Zr, etc. It is formed when Zn+ 2 and a dilute base react or ZnO is 
shaken with H 2 O, and is soluble either in excess strong base forming zincate 
ion, or in ammonium hydroxide, forming tetrammine zinc ion, Zn(NH 3 )J 2 . 
The sp* bond type in the latter leads to tetrahedral geometry. ZnO is 
used in white paints and as a filler in rubber compounding. Zincate- 
strong base solutions are used to plate metals like Al with Zn prior to 
Cu plating. 
Zinc salts are made by dissolving Zn or ZnO in acids. The halides 


are the most water soluble and are also appreciably soluble in organic 
solvents in which they are sometimes employed (as is A1 2 C1 6 ) for catalytic 
effects. ZnCl 2 is used as a flux in soldering because of its ability to dissolve 
metal oxide films. ZnS is precipitated in group 3 by alkaline sulfides and 
is readily soluble in dilute acids. It is prepared on an industrial scale, 
ZnSO 4 + BaS = BaSO 4 + ZnS; the mixture of products is called 
lithopone, a white paint pigment. Other sparingly soluble zinc com- 
pounds besides those in Table 17-6 are the phosphate, Zn 3 (PO 4 ) 2 , and 
ferrocyanide, Zn 2 Fe(CN) 6 , all of which are white. 


Reactions (in order 

of decreasing [Zn +2 ]) 


Formula Weight 

ZnC 2 O 4 = Zn+ 2 + 

^Oa 2 

1.5 x 10- 9 
2 x 10 10 


Zn(OH)o = Zn+ 2 + 
Zn(NH :i )+ 2 = Zn+ 2 

+ 4NH 3 

5 x 10~ 17 
3.4 x 10- ]0 
1 x 10- 20 


ZnSe = Zn+ 2 + Se~ 2 
Zn(CN) 4 2 --= Zn 42 + 4CN~ 

io- 31 

1.2 X IO" 18 


Qualitatively, Zn +2 is detected as the sulfide or ferrocyanide after other 
group 3 metals have been removed. Quantitatively, Zn h2 is often titrated 
with standard Fe(CN)- 4 using UO+ 2 as an external indicator, or it is 
precipitated as Zn(NH 4 )PO 4 and ignited and weighed as Zn 2 P 2 O 7 , zinc 

Group 3 Analysis General Description 

The ions of groups 1 and 2 are first removed as described in the two 
preceding chapters. The solution is buffered with NH 4 OH + NH 4 C1 and 
(NH 4 ) 2 S is added. Tn this alkaline sulfide solution, Al, Cr, and Fe+ 3 
precipitate as hydrous oxides and the other group 3 ions, including Fe+ 2 , 
precipitate as sulfides. Centrifugation of this mixture separates group 3 
in the residue and retains groups 4 and 5 in the centrate. 

The residue is dissolved in nitric acid and separated into two subgroups 
by the addition of excess NaOH. These are called subgroup 3 A (the 
"nickel group") and subgroup 3B (the "aluminum group*'). Subgroup 
3A is a residue of Fe(OH) 3 , Ni(OH) 2 , Co(OH) 2 , and Mn(OH) 2 . Subgroup 
3B is a solution of amphoteric metals as complex hydroxides, A1(OH) 4 , 
Cr(OH) 4 , and Zn(OH) 4 . Adding H 2 O 2 to the mixture does not dissolve 

35 U 



o t is O 










Q, K 



W o 








J ^ ^ 

S^Xffi^ . - 

na !ri xs >^ ^T 05 ^ _ a: _^ _. ^i 

g s 

CA febm 




S3 g 






> 2 
i 8 

PQ C3 


or precipitate any metals but does oxidize* Co(OH) 2 to Co(OH) 3 , Mn(OH) 2 
to MnO 2 , and Cr(OH)j to CrO^ 2 , and the mixture is then centrifuged to 
separate the subgroups. 

Continuing first with subgroup 3A, one dissolves the solids in a mixture 
of hot, dilute HNO 3 and H 2 O 2 to give a solution of Fe+ 3 , Ni+ 2 , Co+ 2 , and 
Mn +2 . The solution is divided into four parts and each tested for a 
different ion. Iron is tested with SCN~, the evidence being the red ferric 
thiocyanate color; nickel is tested with dimethylglyoxime as described in 
Chapter 12 (after precipitating iron and manganese with NH 4 OH), a 
positive test being a bright red precipitate; cobalt is tested with SCN~ 
and acetone (after complexing iron with H 3 PO 4 ), a positive reaction being 
a blue color in the organic layer; and manganese is tested by oxidation 
with HNO 3 + NaBiO 3 to give purple permanganate ion, MnOj\ These 
spot tests are distinctive enough so that a systematic separate scheme for 
subgroup 3A is not necessary. 

Subgroup 3B is acidified with HNO 3 and then made slightly basic with 
NH 4 OH whereupon aluminum precipitates as the hydrous oxide, leaving 
CrO 4 2 and Zn(NH 3 ) 2 in the centrate after centrifuging. Aluminum is 
confirmed by solution and reprecipitation in the presence of aluminon. 
The chromate-tetraammine zinc solution is buffered and chromate 
removed by precipitation with Ba 42 to give yellow, insoluble BaCrO 4 . 
The zinc ions in the centrate are then shown present by precipitation with 
ferrocyanide of white K 2 Zn[Fe(CN) 6 ]. 

This procedure is but one of many workable schemes and the student 
should read the article by Tenbusch and Brewer listed in the references 
at the end of this chapter for others. 


Test solutions contain 10 mg/ml of the metallic ions. 

Test 17-1. Some Reagents for Group 3 Analysis. As time permits, 
try each of the following in separate tests with a few drops of each group 3 test 
soln., including both Fe+ 2 and Fe+ 3 : little NH 4 OH, excess NH 4 OH, little 6 M 
NaOH, excess NaOH, excess NaOH + H 2 O 2 , (NH 4 ) 2 S, KSCN, K 4 [Fe(CN) 6 ], 
and K 3 [Fe(CN) 6 ]. Summarize the results in a table including formulas and 
characteristics of the products. Note which tests might help confirm ions in 
the unknown and what interferences one might encounter in a mixt. of several 
ions. Your data can be of value in seeking confirmatory tests on the unknown. 

Test 17-2. Group 2 and Group 3 Sulfide Precipitations. In solns. 
about 0.3 M in H + , the [S~ 2 ] is sufficient to ppt. group 2 sulfides but not large 

* The potentials in basic solution are such that peroxide is capable of oxidations 
like these. In the next step with acid solution, H 2 O 2 acts as a reducing agent, however. 


enough to ppt. group 3 sulfides, nor is [OH~] large enough to ppt. group 3 
hydroxides. On this basis is the separation between these groups effected. 

To illustrate, put 5 drops of Cd+ 2 soln., 5 drops of Fe+ 2 soln., 5 drops 
of H 2 O, 1 drop of 6 M HC1, and 5 drops of thioacetamide in a test tube and heat 
it in a water bath. Only yellow CdS ppts. Consult the K sr values and see 
why this is expected. Add 2 drops of 6 M NaOH to neutralize the acid and 
increase the [S~ 2 ]. What is the black ppt. now forming? 

Test 17-3. Separation of a 3A 3B Mixture. Make a mixt. of 4 drops 
of A1+ 3 test soln., 4 drops of Fe+ 3 , and 10 drops of H 2 O. Review the behavior 
of these ions with excess strong base as noted in Test 1 7-1 . Add 3 drops of 
6 M NaOH to the mixt. and stir. What is the residue and what is in soln. ? 
Centrifuge, remove the centrate, and neutralize it with 2 M HAc. What is the 
ppt. that forms? Write the equations and explain how this method effected 
the separation (later used in the unknown procedure) between typical ions of 
the subgroups. 

Test 17-4. Conversion of Cr(OH) 3 to Cr(OH)^ CrO^ 2 , Cr 2 Of 2 and 
OO 5 . These transformations are described under Chromium in this chapter 
and in part H, Chapter 22. Try the reactions and account for the color changes 
by writing equations. Start by preparing some chromic hydroxide and carry 
the sample through the operations indicated. 

Test 17-5. Oxidation States of Mn. 

(a) Mn+ 2 to MnO. Mix in a crucible 2 drops of Mn +2 test soln., 10 drops 
of H 2 O, 2 drops of coned. H ;J PO 4 , 6 drops of dil. HNO 3 , and warm. Now 
add a few mg of one of the following oxidizers: NaBiO 3 , sodium bismuthate; 
(NH 4 ) 2 S 2 O 8 , ammonium peroxydisulfate; Na 3 H 2 IO 6 , trisodium paraperiodate. 
Warm the mixt. again. Note the red color of permanganate against the white 
background. The concn. limit is 0.1 ppm Mn. 

(b) Mn +2 to Mn0 2 . Mix 4 drops of Mn +2 soln., 10 drops of H 2 O, and 2 drops 
of 6 M NaOH. The ppt. is Mn(OH) 2 and it will gradually turn dark due to 
atmospheric oxidation to hydrated manganese dioxide, sometimes written 
MnO(OH) 2 . This can be hastened by adding 6 drops of 3 % H 2 O 2 . Centrifuge 
and save the residue for (c). 

(c) Mn0 2 to MnOt*. This is our first bead test and the instructor will 
demonstrate it first. Prepare a carbonate bead by fusing some Na 2 CO 3 on a 
hot Pt wire. (Nichrome wire is not recommended as a substitute in this parti- 
cular test.) Reheat the bead, touch it to the MnO 2 prepd. in (6), and reheat. 
Now touch it to a crystal of KC1O 3 and reheat. The green color is due to the 
manganate ion, MnO^ 2 . 

As an alternate confirmation of the MnO 2 ppt., one can convert it to MnO,f 
by the combination of reagents in (a). 

Test 17-6. Aluminon Test. Aluminon is a common name for an org. 
reagent used to colorimetrically detect aluminum in concns. lower than can be 
visibly pptd., as A1(OH) 3 . Dilute 1 drop of A1+ 3 test soln. to 10ml and put 
2 ml of that in a test tube. Add a drop of 1 M HAc, 4 drops of NH 4 Ac, 3 drops 


of aluminon, and mix. Compare the color to that of a blank made by using 
simply 2 ml of dist. water with buffer and reagent. See Example 16, Chapter 
12, p. 197. Fe+ 3 and Cr +3 interfere with this test by giving similar reactions. 

Test 17-7. Thenard's Test for Al. This is our first blowpipe test and 
the instructor will demonstrate it first. Put a spatula full of A1(OH) 3 , which 
you have prepd., on a charcoal block, add a drop of 0.05 M Co+ 2 soln., and 
direct a Bunsen flame to the mixt. with a blowpipe. A delicate blue (Thenard's 
blue) in the well-ignited residue is indicative of Al, due probably to the presence 
of cobalt aluminate, Co(AlO 2 ) 2 . 

Alternately the test can be run on either some rolled up filter paper or on 
asbestos fiber held on a Pt or nichrome wire. The matrix is impregnated with 
the test chemicals and ignited in the Bunsen flame directly. Under the same 
conditions, Zn+ 2 gives a green color and Mg+ 2 a pink color. A blank should 
also be run since Co f 2 gives a little color of its own. 

Test 17-8. Cobalt Bead Test. See part 1, special experiment 4. Pre- 
pare first a small sample of CoS, then a borax bead. Fuse a little of the sulfide 
on the bead and notice the color produced after several minutes heating. It 
may be due to cobalt metaborate, Co(BO 2 ) 2 , or the complex salt Na 2 [Co(BO 2 ) 4 ]. 

Test 17 9. K 3 Co(NO 2 ) 6 . To 3 drops of Co+ 2 test soln., add 5 drops of 
H 2 O, a drop of 6 M HAc, 3 drops of KNO 2 , and warm the tube in a water bath. 
The yellow, slowly forming ppt. is potassium hexanitrocobaltate(III). This is 
a good test for cobalt in the presence of nickel. 

Test 17-10. Co+ 2 ~Acetone Test, (a) Make a dil. Co +2 -Fe+ 3 mixt. Add a 
drop of KSCN and note the red color of the ferric-thiocyanate ions. Decolorize 
to a light pink by dropwise addns. of 6 M H 3 PO 4 and good mixing, then add 
3 drops in excess. Fe +3 is complexed (see test 22-17, part b) by this means. 
Now add another drop of KSCN, then carefully put 5 drops of acetone, 
(CH 3 ) 2 C= O, in the tube to form a layer on top. The blue color, possibly 
Co(SCN) 2 [O=-C(CH 3 ) 2 ] 2 ~ 2 , proves the presence of cobalt. The concn. limit is 

(b) The student may optionally wish to try variations of the test using for 
example sodium potassium tartrate to complex the iron and a higher ketone 
like methyl isobutyl ketone to coordinate cobalt. 

Test 17-11. Co+ 2 and Ni+ 2 Compared, Using Dimethylglyoxime. 

Dilute 4 drops of Co +2 test soln. to 1 ml in a tube, and in another tube prepare a 
similar diln. of Ni+ 2 . To each add 3 drops of 0.5 M NH 4 C1, a drop of 3 M 
NH 4 OH, and 3 drops of dimethylglyoxime. Compare the results with reference 
to example 1, Chapter 12, p. 193 for the equation. The concn. limit is about 
3 ppm Ni. Fe+ 2 gives a red color with the reagent and should be oxidized 
to Fe+ 3 with H 2 O 2 if present. 

Test 17-12. Co+ 2 and Ni+ 2 Compared, Using oL-Nitroso-fi-naphthol. 

Make dilns. as in test 17-11. To each add a drop of 6 M HAc and 4 drops of 


organic reagent and heat the tubes in a water bath. Note results and see 
paragraph 6, Chapter 12, for equation. 

Test 17-13. Rinmann's Test for Zn. Repeat test 17-7 using a little 
ZnS or Zn(OH) 2 in place of A1(OH) 3 . The color produced is probably due to 
CoZnO 2 and is called Rinmann's green. 

Test 17-14. Dithizone Test for Zn 12 . See special experiment 10, 
Chapter 23, and example 2, Chapter 12. 

Analysis of Known Mixtures 

A known group 3 sample may be analyzed prior to analysis of the 
unknown, or several simple mixtures as in test 17-3 may be tried to see if 
one has in mind the separations needed, as well as short cuts, in which 
some ions may be tested directly in the presence of others. If no further 
preliminary study of this nature is undertaken, the student, having his 
notebook writeup completed, proceeds to the unknown sample and 
follows the procedure for group 3 analysis. Some of the confirmatory 
and spot tests from the preliminary tests may be used also at his discretion. 

Interfering Ions 

Of the unions discussed in Chapter 22, BOj, F~, C 2 Oj" 2 , and PO 4 3 
hinder analysis by complexing some group 3 cations and/or prematurely 
precipitating group 4 cations. Directions are given in Chapter 21 for 
circumventing these difficulties but they are not present unless so stated 
by the instructor. 


A. Reactions with NH^Cl + NH^OH 

(1) Fe+ 3 + 3NH 4 OH (N = 4t> Fe(OH),, /<r + 3NHJ 

(2) Cr*- 3 + 3NH 4 OH (N = ' } Cr(OH) 3 , r + 3NH+ 

(3) A1+ 3 + 3NH 4 OH (N = f) Al(OH) 3|r + 3NH+ 

(4) Co+ 8 + 6NH 4 OH (N = ^ CoCNHjtf 2 + 6H 2 O 
4Co(NH 3 )J 2 + 2H 2 + 2 (air) = 4Co(NH 3 ) + 4OH~ 

(5) Ni+ 2 + 6NH 4 OH (N = l> Ni(NH 3 ) 6 ^ y + 6H 2 O 

(6) Fe+ 2 + 2NH 4 OH <N = 4<) Fe(OH) 2(;r + 2NHJ 
4Fe(OH) 2 + 2H 2 O + 2 (air) = 4Fe(OH) 3 


(7) Zn^ 2 + 4NH 4 OH = 4 Zn(NH 3 )+ 2 + 4H 2 O 

(8) Mn+ 2 + NH 4 OH (N = +) No reaction 

B. Reaction of the Above upon (NH^) 2 S Addition 

(9) 2Fe(OH) 3 + 6NH+ + 3S~ 2 = 2FeS* + 6NH 4 OH + S r (some) 

2Fe+ 3 + 3S~ 2 = Fe 2 S 3/? (also in cold solution) 
Fe 2 S 3 + 6H 2 = Fe(OH) 3 + 3H 2 S (when heated) 

(10) Cr(OH) 3 + 2NH+ + S~ 2 = No reaction 

(11) A1(OH) 3 + 2NH+ + S- 2 = No reaction 

(12) 2Co(NH 3 )+ 3 + 3S~ 2 = 2CoS^ + 12NH 3 + S F 

(13) Ni(NH 3 ) 2 + S- 2 = NiS 7/ + 6NH 3 

(14) Zn(NH 3 )J 2 + S- 2 = ZnS> v + 4NH 3 

(15) Mn^ 2 + S~ 2 = MnS^ 

C. Acid Action on Sulfides and Hydroxides 

(16) Fe(OH) 3 + 3H^ = Fe+ 3 + 3H 2 O 

FeS + 4H^ + NO 3 = Fe+ 3 + NO + S + 2H 2 O 

(17) Cr(OH) 3 + 3H^ = Cr t3 + 3H 2 O 

(18) A1(OH) 3 + 3H = A1+ 8 + 3H 2 

(19) 3CoS + 8H+ + 2N0 3 = 3Co +2 + 2NO + 3S + 4H 2 O 

(20) 3NiS + 8H+ + 2NO 3 = 3Ni+ 2 + 2NO +_3S + 4H 2 O 

(21) 3ZnS + 8H+ + 2NO 3 = 3Zn+ 2 + 2NO + 3S + 4H 2 O 

D. Reaction with Excess NaOH 

(22) Fe+ 3 + 30H- = Fe(OH) 3 

(23) Cr+ 3 + 40H- = Cr(OH) 4 

(24) A1+ 3 + 4OH- = A1(OH) 4 

(25) Co+ 2 + 20H- = Co(OH) 2/?u = Co(OH) 2p 

(26) Ni+ 2 + 20H- = Ni(OH) 26> 

(27) Fe+ 2 + 20H- = Fe(OH) 2f;r 

(28) Zn+ 2 + 4OH- = Zn^H)^ 2 

(29) Mn+ 2 + 20H- = Mn(OH) 2/> _ w 


E. Reaction of the Above with H 2 O 2 in Basic Solution 

(30) Fe(OH) 3 + H 2 2 = No reaction 

(31) 2Cr(OH)j + 3H 2 2 + 2OH~ = 2CrO^ + 8H 2 O 

(32) A1(OH)4 + H 2 2 = No reaction 

(33) 2Co(OH) 2 + H 2 O 2 = 2Co(OH) 3i? 

(34) Ni(OH) 2 + H 2 O 2 = No reaction 

(35) 2Fe(OH) 2 + H 2 O 2 = 2Fe(OH) 3 

(36) Zn(OH)j 2 + H 2 2 == No reaction 

(37) Mn(OH) 2 + H 2 O 2 = MnO 2i , + 2H 2 O 

F. Subgroup 3A Resolution in HNO 3 + H 2 O 2 

(38) Fe(OH) 3 + 3H+ = Fe^ 3 + 3H 2 O 

(39) 2Co(OH) 3 + H 2 2 + 4H' = 2Co+ 2 + 6H 2 O + O 2 

(40) Ni(OH) 2 + 2H+ = Ni+ 2 + 2H 2 O 

(41) Mn0 2 + H 2 2 + 2H+ = Mn+ 2 + 2H 2 O + 2 

G. Subgroup 3B with Excess HNO^ 

(42) 2CrO4 2 + 2H+ = Cr 2 O^ 2 + H 2 O 

(43) A1(OH)4 + 4H+ = A1+ 3 + 4H 2 O 

(44) Zn(OH)j 2 + 4H+ = Zn+ 2 + 4H 2 O 

H. Iron 

(45) Fe^ 3 + SCN- = FeSCNJ 2 

(46) 4Fe^ 3 + 3Fe(CN)~ 4 = Fe 4 [Fe(CN) 6 ] 3fltt 

/. Cobalt 

O O 


(47) Co+ 2 + 2SCN- + 2CH 3 C CH 3 = Co(SCN) 2 (CH 3 

(48) Co+ 2 + 3K+ + 7NO2- + 2HAc 

= K 3 [Co(N0 2 ) 6 ] y + NO + H 2 

(See also paragraph 6, Chapter 12, p. 194.) 

J. Nickel 

No additional reactions. (See also example 1, Chapter 12.) 


1C. Manganese 

(49) 2Mn+ 2 + 14H+ + SBiOg = 2MnO 4 ; u + 5Bi+ 3 + 7H 2 O 

(50) 2MnO 2 + 10H+ + 3BiO~ = 2MnO 4 + 3Bi+ 3 + 5H 2 O 

L. Chromium 

(51) Cr(OH) 4 + H+ = Cr(OH) 3 + H 2 O 

(52) Ba+ 2 + Cr0 4 2 = BaCrO 4r 
(See also part H, Chapter 22.) 

M. Aluminum 

(53) AKOH)^ + H+ = A1(OH) 3 + H 2 O 
(See also paragraph 16, Chapter 12.) 

N. Zinc 

(54) 2K+ + Zn+ 2 + Fe(CN) 6 4 = K 2 Zn[Fe(CN) 6 ]^ 

(See also example 2, Chapter 12.) 


04) If using a solution known to contain only the metals of group 3, 
dilute 1 ml of sample with 2 ml of H 2 O. Add 8 drops of 2 M NH 4 C1, 
and put the tube in a hot-water bath. After several minutes add 15 M 
NH 4 OH dropwise with good stirring, and continue addition until the 
solution becomes distinctly alkaline as evidenced by a red color with 
phenolphthalein when a drop of solution is transferred with the stirring 
rod to a drop of indicator on the spot plate. Now add 8 to 10 drops of 
(NH 4 ) 2 S reagent, continue to stir, and keep warm for another 2 to 3 
minutes; then rinse off the stirring rod and centrifuge. The centrate is 
discarded and the residue retained for group 3 analysis, (C) below. 

If at this point in the general unknown analysis, phosphate and borate 
require removal according to directions given in Chapter 21, the instructor 
will so indicate. 

(B) If the sample is a general unknown, the centrate from paragraph (C), 
Chapter 1 6, is the solution used. As this already contains some ammonium 
acetate and unhydrolyzed thioacetamide, the procedure will be slightly 
different from (A), above. Heat the solution in a water bath, add 4 drops 
of NH 4 C1, and then, with stirring, 15 M NH 4 OH until alkaline. After a 
few minutes, add 2 drops of (NH 4 ) 2 S, and see if any further precipitation 


takes place. If so, add drops of sulfide reagent until precipitation is 
complete. Wash down the stirring rod and tube walls with a few drops of 
H 2 O, centrifuge, and remove the centrate and save it for groups 4 and 5. 
Wash the residue with a hot mixture of 1 ml of H 2 O and 4 drops of NH 4 C1, 
centrifuge, and discard the wash liquid. The residue is analyzed starting 
with (C). 

(C) The residue from (A) or (B) may consist of any mixture of Fe(OH) 3 , 
Cr(OH) 3 , A1(OH) 3 , CoS, NiS, FeS, ZnS, and/or MnS. Add to it 1 ml of 
6 M HNO 3 , and stir while warming in the hot-water bath. If the precipi- 
tate is not dissolved, except for a little sulfur,* which usually aggregates 
and floats as a small spongy mass, add another 10 drops of acid and 
continue heating. Remove the sulfur and rinse the solution into a 20-ml 
beaker with the aid of a few drops of H 2 O. Evaporate the solution slowly 
under the hood to a volume of about 6-8 drops. Avoid splattering and 
avoid evaporation to dryness. Rinse the solution into a test tube with 
the aid of several i ml portions of water. Heat in a water bath and 
add drops of 6 M NaOH until alkaline, then 8 drops more. Finally, put in 
10 to 12 drops of 3% H 2 O 2 , mix, and continue to keep the tube hotj for 
4 to 5 minute , K nger. Centrifuge. Put the centrate in a 20-ml beaker 
and save for ( E\. Wash the residue 2 or 3 times with about 1 ml of hot H 2 O 
each time, combining the first wash extract with the centrate and discarding 
subsequent washings. Analyze the residue according to the next 

(D) The residue from (C) is group 3 A and may contain any combination of 
Fe(OH) 3 , Ni(OH) 2 , Co(OH) 3 , and MnO 2 . With 30 to 40 drops of 6 M 
HNO 3 rinse the mixture into a 20-ml beaker. Heat carefully with a low 
flame, then add 4 to 6 drops of 3 % H 2 O 2 and boil the solution carefully. 
A little more HNO 3 and H 2 O 2 may be added if the residue does not dissolve 
completely in a few minutes. Let the solution cool, add 2 ml of water 
and divide the solution among 4 tubes: 

Tube 1. Add 3 drops of SCN - solution. A dark red color as in 
test 17-4 shows iron is present. A faint pink is reported as a trace of Fe. 

Tube 2. If iron was present, completely precipitate it with 15 M 
NH 3 , centrifuge, and just neutralize the centrate with 6 M HAc. Add 
6 drops of dimethylglyoxime to this solution and a drop of 6 M NH 4 OH 
and mix. A bright red precipitate proves nickel was in the sample. 

* As in nitric acid solutions of group 2 sulfides, small amounts of sulfides may be 
occluded in the sulfur residue and color it darkly. The loss will be small and is neglected 

t Continued heating not only helps coagulation but also hastens decomposition of 
excess H 2 O 2 , which would reduce CrO~ 2 in (E) when acidified. In basic solution hydrogen 
peroxide oxidizes Cr m to Cr VI , but in acidic solution the reverse occurs. 


Review test 17-12. If iron was not in Tube 1, neutralize with HAc and 
NH 4 OH and proceed as above with the organic reagent, followed by a 
drop of NH 4 OH. 

Tube 3. If iron was present, add 3 drops of SCN~ and as in test 17-10 
decolorize the iron thiocyanate with excess 6 M H 3 PO 4 . Add another 
drop of SCN~ solution, then without shaking the mixture add 5 drops of 
acetone. A blue organic layer is proof that cobalt is present. 

If iron was not found in Tube 1, H 3 PO 4 is omitted, and SCN~ and 
acetone are added initially. The blue color when Co +2 is present is the 

Tube 4. Add 3 drops of concentrated H 3 PO 4 , 3 of concentrated HNO 3 , 
heat, and add a bit of solid NaBiO 3 . A red to purple color as in test 17-5 
shows manganese is present. 

This is the end of the 3 A procedure. 

(E) The centrate from (C) may contain any combination of A1(OH) 4 , 
Zn(OH) 4 2 , and/or CrO 4 2 . Boil the mixture for a minute. Then add 
concentrated HNO 3 until the mixture is acidic, transfer it to a test tube, 
and add concentrated NH 4 OH until it is strongly basic again. A yellow 
color indicates CrO~ 2 is present, and a white flocculant precipitate is 
A1(OH) 3 .* Centrifuge and remove the centrate to another tube for 
analysis in paragraph (G). Save the residue for (F). 

(F) Wash the residue with 2 ml of hot H 2 O and discard the washings. 
Redissolve the residue in a few drops of 6 M HAc, add 1 ml of H 2 O, 
4 drops of aluminon, and heat in the water bath. When hot, add a few 
drops of 3 M NH 4 OH to make it neutral or barely basic. A red precipitate 
confirms aluminum as in test 17-6. 

(G) The centrate from (E) contains CrO 4 2 only if it is yellow but may also 
contain colorless Zn(NH 3 )J 2 . If it is colorless, neutralize it with 6 M 
HAc and add 2 drops excess; then put in 6 drops of K 4 Fe(CN) 6 . A 
whitish precipitate shows zinc is present as in test 17-1. 

If the centrate from E is yellow, just neutralize it with 6 M HAc. Now 
put in drops of BaCl 2 until precipitation of BaCrO 4 is complete and 
centrifugation should yield a colorless centrate. Carry the centrate to 

(H) The yellow residue shows the presence of chromium in the sample. 
This element may be confirmed according to test 22-21. 

(H) Treat the centrate with 2 drops of 6 M HAc and 6 drops of 
K 4 Fe(CN) 6 . As before, a whitish residue indicates zinc. 

This is the end of the group IB procedure. 

* A trace of gelatinous silicic acid from the dissolution of glass by alkali is usually 
visible here. Al+ 8 from the sample should give a strong test. Silicic acid gives only a 
white color in the aluminon test. 



1. One has a greenish suspension of Fe(OH) 2 in water at pH 1. Air is 
bubbled through and after a while the suspension is noted to have turned brown 
and the solution pH dropped to 6. Explain with equations. 

2. There is a generalization that in the ferrous and ferric compounds formed 
with ferro- and ferricyanide, dark color is only noted in the compounds con- 
taining iron of both valences. What experimental evidence can you offer here ? 

3. Write equations to show how one could start with chromite ore and a 
few common chemicals and prepare two industrially important products: 
potassium dichromate and Prussian blue. 

4. Outline a separation of Fe+ 3 and Cr+ 3 based on the amphoterism of the 

5. Show by calculation whether or not FeS can precipitate from a hot mixture 
of 1 drop of 5% CH 3 CSNH 2 , 2 drops of 6 M NaOH, and 1 drop of 0.01 M 
FeSO 4 in a total volume of 2 ml. 

6. Complete: 

OH- H^ 

(a) ^ Cr(OH) 3 ^ 

H+ OH- 

(b) Cr 2 0f 2 + H+ + H 2 2 = 

(c) Na 2 2 + Cr(OH)J- + H 2 O = 

7. J. P. Slipshod proposes the following group 3 shortcuts for analysts in a 

(a) "If an unknown may contain only Zn +2 and Fe+ 3 , addition of Fe(CN)^" 4 
will give a white precipitate with zinc. After centrifuging it out, the centrate 
gives a red color for iron with SCN~. 

(b) "In a Co +2 -Ni+ 2 mixture, a borax bead test is run on a portion, giving a 
grey color for Ni+ 2 , and to the other portion is added (NH 4 ) 2 S, giving a black 
precipitate of CoS which shows cobalt is present. 

(c) "In a Cr +3 -Al+ 3 mixture, add hot NaOH to solubilize amphoteric aluminum 
and leave Cr(OH) 3 . The centrate is mixed with aluminon giving a red color, 
and the residue is dissolved in H 2 SO 4 which gives a blue color of CrO 5 in butyl 

Mention wherein these procedures are open to question. 

8. How would you test a metal foil to see if it is tin or aluminum? 

9. HNO 3 , but not HC1, may be shipped in aluminum tank cars. Explain. 

10. Many other group 3 procedures are used besides the one given in this 
chapter. One precipitates the group as done here, then dissolves all but NiS 
and CoS in dilute HC1. The centrate from that separation is treated with 
concentrated NH 4 OH in which Zn+ 2 and Cr+ 3 are complexed and are soluble. 
The hydroxide residue is separated and treated with NaOH in which A1(OH) 3 

With regard to the fractions separated, the CoS-NiS mixture is dissolved in 
more concentrated, hot HC1, and then neutralized and analyzed with the organic 


reagents. The Zn-Cr ammonia complexes are reacted with Na 2 O 2 to give 
Zn(OH)i~ 2 and CrO^ 2 , the latter being precipitated with Ba +2 , the former just 
acidified and precipitated with ferrocyanide. Draw a flow sheet for this and 
include iron and manganese in the scheme as they would react, and include a 
separation and confirmation of them. 

11. Using all the information you have on group 3, propose a separation 
scheme of your own invention for those seven ions and present it as a flow sheet. 

1 2. (Library) Copy or prepare a flow sheet on group 3 from another qualitative 
text and compare it with the one in this book. Try to criticize one point in 
each as being unsatisfactory by virtue of its technical difficulty, possible incom- 
plete separation, or other reason. 

1 3. Give equations for application of the Goldschmidt process to chromite ore. 

14. A rust inhibitor for radiators is supposed to contain potassium chromate. 
What action does this compound have on iron and how would one test for CrO^ 2 
in a mixture of simple salts? How could one tell by inspecting the radiator 
drain water that iron corrosion had been going on ? 

15. Aluminum is chemically quite active and the value for Al = A1+ 3 -f 3e~ 
is high. Why don't airplanes made with a great deal of almost pure Al skin 
corrode badly in moist air? Why does pure Al corrode less readily than its 
alloys? Trimethyl aluminum, (CH 3 ) 3 A1, has been used as a rocket fuel. Cite 
one advantage of it. How might it be synthesized? If burned with liquid O 2 
what are the probable products ? 

16. (a) A group 3 unknown is suspected to contain only Mn +2 and Co f2 . 
What quick methods are available for testing the suspicion? 

(b) Repeat (a) for a sample suspected to contain only Fe +3 and Zn 1 2 . 

(c) Repeat for a sample thought to contain Ni+ 2 , Cr f3 , and A1+ 3 . 

17. There is a little aluminum in some glass, and strong basic solutions stored 
in glass are known to dissolve it slowly. How could one analyze a sample of 
6 M NaOH for Al? In what form would the aluminum be? 

18. What group 3 ions may be present in each of the following: 

(a) A solution known to contain cations of group 3 only is colorless. 
(6) A group 3 unknown gives a blackish precipitate with (NH 4 ) 2 S and when 
that is dissolved in hot HC1, a light green solution results. 

(c) A group 3 solution gives a black precipitate and yellow supernatant 
liquid when reacted with NaOH + H 2 O 2 . The precipitate is soluble in H 2 SO 4 
-}- H 2 O 2 to yield a solution which has a faint pink tint. 

(d) A mineral is a black powder. It is dissolved in HC1 and when excess 
NH 4 OH is added a brown flocculent precipitate comes down. 

(e) What color would one observe after dissolving a piece of 18-8 stainless 
steel in HC1 ? What would be the result of adding excess NaOH to that solution ? 

19. Co(OH) 3 is formed in basic solution by air oxidation of Co(OH) 2 . When 
the mixture is acidified with HC1, the odor of chlorine gas is detectable. Explain. 

20. A solution of Cr+ 3 is acidified with H 2 SO 4 and a piece of Zn is added. 
The color gradually changes from dark green to blue. Explain, and include 
oxidation potential data for support. 

21. Draw a flowsheet for the preparation of a pig of bronze (Cu-Zn-Sn) 


starting with chalcopyrite, zincite, and cassiterite ores and any other materials 

22. Dimethylglyoxime will give a color with as little as 3 y of Ni. If 10 mg 
of alloy are dissolved and tested with the reagent, what is the minimum per- 
centage by weight Ni in the sample that can be detected? (Ans. 0.03%) 


1. D. C. Schooley, J. Chem. Educ., 23, 399 and 522 (1946). (Al, Co) 

2. R. K. McAlpine, J. Chem. Educ., 23, 298 (1946). (Co, Ni, Zn, Mn) 

3. M. M. Tenbusch and G. E. F. Brewer, /. Chem. Educ., 23, 66 (1946). (Group 3) 

4. J. H. Parsons, J. Chem. Educ., 25, 207 (1948). (Color in transition elements) 

5. W. E. Morrell, /. Chem. Educ., 27, 274 (1950). (Zn) 

6. B. J. Lerner, C. S. Grove, and R. S. Casey, /. Chem. Educ., 29, 438 (1952). (Fe 

7. S. B. Smith and J. M. Shute, J. Chem. Educ., 32, 380 (1955). (Al separation) 

8. J. Kleinberg, J. Chem. Educ., 33, 73 (1956). (Unfamiliar oxidation states) 



Magnesium may be analytically segregated with this group or with 
group 5 depending upon the group 4 precipitating conditions (illustrated 
in Chapter 8 by calculations). In the methods described here, Mg+ 2 is 
put with group 5 and discussed there despite its obvious relationship to 

other "alkaline earth" metals of periodic 
group IIA. Beryllium chemistry is described 
and illustrated with experiments in Chapter 
20. Radium is not dealt with here, but 
other than its rarity and radioactivity it is 
quite similar to Ba. 




















FIG. 18-1. The periodic 

table in the vicinity of the 

group 4 metals. 


Barium is a soft, silvery metal of very 
limited use. Its sp. gr. is 3.75, its m.p. is 
717 C, and b.p. is 1640 C. The main 
mineral is barites, BaSO 4 . A mixture of 
BaSO 4 and ZnS is called lithopone and is the 
basis for some white paints. 

While a subchloride and subhydride of 
barium, BaCl and BaH, have been reported, 
the only stable compounds are those in which 
the oxidation state is (II). 




Barium has the same order of high chemical activity as the alkali 

Ba = Ba+ 2 + 2e~ E = 2.90 

It easily decomposes water to yield H 2 + Ba(OH) 2 . The metal tarnishes 
rapidly in air and forms at elevated temperatures a hydride with H 2 , 
BaH 2 ; an oxide and peroxide with O 2 , BaO and BaO 2 ; and a nitride 
with N 2 , Ba 3 N 2 . These compounds react with water to yield, in order, 
Ba(OH) 2 + H 2 ; Ba(OH) 2 ; Ba(OH) 2 + O 2 ; Ba(OH) 2 + NH 3 . Barium 
hydroxide is a stronger base than either Sr(OH) 2 or Ca(OH) 2 and is the 
most difficult to dehydrate to the oxide. The peroxide is also the most 
stable of those formed by the alkaline earths, as is typical of other barium 

Among other compounds of barium, the acetate, bromide, chloride, 
chlorate, iodide, and nitrite are quite soluble; the fluoride, hydroxide, and 
nitrate somewhat less soluble; and the salts listed in Table 18-1 are very 
slightly soluble. Barium sulfide is not formed in water solution since the 
solid reacts with H 2 O giving H 2 S + Ba(HS) 2 + Ba(OH) 2 . As typical of 
large metallic ions, Ba+ 2 forms complexes weakly. 


Reactions (in order of decreasing [Ba* 2 ]) 


Formula Weight 

Ba(OH) 2 8H 2 O = Ba+ 2 + 2OH~ + 8H 2 O 

5.0 x 10- 3 


BaF 2 = Ba+ 2 + 2F~ 

2.4 x 10~ 5 


BaC 2 O 4 2H 2 O = Ba+ 2 + C 2 <V + 2H 2 O 

1.5 x 10- 8 


BaC0 3 = Ba+ 2 + COgf 2 

1.6 x 10 9 


BaSO 4 = Ba+ 2 + SO^ 2 

1.5 x 10 ~ 9 


BaSO 3 = Ba+ 2 + SO^ 2 

9.5 x 10- 10 


BaCr0 4 ==Ba+ 2 + Cr(V 

8.5 x 10- 11 


BaSeO 4 = Ba+ 2 + SeO^ 2 

2.8 x 10 1J 


Ba 3 (P0 4 ) 2 = 3Ba+ 2 -f 2PO 4 3 

6 x 10 - 39 


Quantitatively, Ba 42 is determined by weighing it in the form of BaSO 4 . 
One qualitative test is the yellow-green flame and another is the fact that 
BaCrO 4 is much less soluble than other group 4 chromates. 


This metal, like the other alkaline earths, is silvery white but harder and 
more brittle than Ba. It has no use as an engineering metal. Its sp. gr. 
is 2.6, the m.p. is 771 C, and the b.p. is 1384 C. The principal minerals 



are strontionite, SrCO 3 , and celestite, SrSO 4 . Sr(NO 3 ) 2 and Sr(ClO 3 ) 2 
are used to impart a red flame color in fireworks and Sr(OH) 2 is employed 
in certain sugar refining processes, but uses of strontium compounds are 
generally quite limited. 

There are no stable Sr 1 compounds so we may say that the oxidation 
state of this element is (II) only. It decomposes H 2 O giving Sr(OH) 2 + H 2 , 
its powerful reducing character obvious from the high positive potential : 

Sr = Sr+ 2 + 2e~ E = 2.89 

The metal burns brightly in air to give only the oxide SrO, which reacts 
with H 2 O to form the strong base Sr(OH) 2 whose saturated water solution, 
however, is only 0.074 M at 25 C. The metal also forms a hydride SrH 2 
with H 2 and nitride Sr 3 N 2 with N 2 at higher temperatures. 

The same solubility generalizations given for Ba+ 2 compounds hold 
for those of Sr+ 2 . The chromate is more soluble in the case of Sr, but it 
can be precipitated if alcohol is added to the water solution, a procedure 
used in the unknown sample manipulation described later. Strontium 
compounds impart a bright red color to the bunsen flame (see special 
experiment 5), which is useful as a qualitative test. 


Reactions (in order of decreasing [Sr+ 2 ]) 


Formula Weight 

Sr(OH) 2 8H 2 O = Sr+ 2 + 2OH~ + 8H 2 O 

3.2 x 10~ 4 


Sr(HPO 4 ) = Sr+ 2 -+ HPO^ 2 

2 x 10~ 4 


Sr(HC0 3 ) 2 = Sr+ 2 + 2HCCV 

1.83 x 10- 6 


SrCr0 4 = Sr+ 2 + CrO^ 2 

3.6 x 10~ 5 


SrSO 4 = Sr+ 2 + SO 4 2 

7.6 x 10~ 7 


SrF 2 = Sr+ 2 + 2F~ 

7.9 x 10~ 10 


SrC 2 O 4 H 2 O = Sr+ 2 + C 2 (V + H 2 O 

5.61 x 10- 8 


SrC0 3 = Sr+ 2 + C<V 

7 x 10~ 10 


Sr 3 (P0 4 ) 2 = 3Sr+ 2 + 2PO4- 3 

I x 10- 31 


Sr 3 (AsO 4 ) 2 = 3Sr+ 2 + 2AsO 4 - 3 

2 x 10- 49 


Quantitatively, Sr+ 2 is precipitated and weighed as SrSO 4 after Ba+ 2 is 
removed as the chromate and Ca H 2 as the oxalate. 


Calcium is silvery and brittle. Its sp. gr. is 1.55, its m.p. is 851 C, 
and b.p. is 1487 C. As a metal, it is used in lead alloys for electric cable 
sheathing and as a degasifier in steel making. Common minerals are 


dolemite, CaCO 3 MgCO 3 , gypsum, CaSO 4 2H 2 O, apatite, (CaF)Ca 4 (PO 4 ) 3 
Cl, aragonite and calcite, two forms of CaCO 3 , and fluorite, CaF 2 . Ca 
is the third most abundant metal in the earth's crust. It is obtained by 
methods typical of the alkaline earths, namely electrolysis of its molten 
chloride-fluoride mixture in a graphite crucible (anode), using an iron 

The only stable oxidation state is (II), though some Ca r compounds 
have been formed under special conditions. The metal is very active 
chemically as a reducing agent: 

Ca = Ca+ 2 + 2e~ E = 2.87 

and it also combines with H 2 , O 2 , and N 2 when heated in those atmospheres 
yielding CaH 2 , CaO, and Ca 3 N 2 , respectively. 

Ca+ 2 is found in natural brines and waters, as well as in many rocks, 
and with Mg+ 2 constitutes the "hardness" in the water. These ions react 
with conventional soaps, which are sodium and potassium salts of long- 
chain aliphatic acids, to precipitate the calcium and magnesium soaps: 

2R C< + Ca+* = R Of Ca + 2Na+ 

X)Na \ x O/ 2 

This process softens the water but of course is not the best way to do so. 
The softening of water is of prime importance particularly in heat exchange 
equipment since calcium salts may precipitate by chemical reaction 

Ca(HCO 3 ) 2 = CaCO 3 + H 2 O + CO 2 

or simply by evaporation at the container walls. These deposits insulate 
boiler tubes, resulting in burn outs, since higher heats are needed to 
maintain normal operation. Calcium bicarbonate in the reaction above 
is known as "temporary hardness" since water bearing it may be partially 
softened by heating. Other calcium salts are not removed in similar 
fashion. Two additional softening methods are worthy of mention: 
first, the zeolite process in which a complex silicate-aluminate exchanges 
2Na + for each Ca+ 2 and can be regenerated for further softening by 
utilizing mass action with a strong brine (NaCl) solution; and second, the 
general process in which softening chemicals are added to the water, such 
as borates, silicates, and phosphates which are capable of forming soluble 
complexes with Ca+ 2 . 

There are quite a few important calcium compounds. The carbonate 
is calcined to give lime, CaO, which in a water slurry is called slaked lime, 
Ca(OH) 2 . Saturated Ca(QH) 2 is 0.021 M at 25 C. Calcium carbonate is 
an important raw material also in the cement and glass industries and as a 
fluxing material in several metallurgical processes, including iron smelting. 



The CO 2 obtained from roasting CaCO 3 is used with NH 3 and NaCl in the 
Solvay process for the production of Na 2 CO 3 and NaHCO 3 . The residue 
of CaO is slaked and used to recover NH 3 from the Solvay by-product 
NH 4 C1. Calcium chloride produced from that step finds use (a) on gravel 
roads where it keeps down dust by removing moisture from the air (mono, 
di, tetra, and hexa hydrates are known) (b) as a refrigeration plant brine, 
and (c) as a deicing salt for winter roads. Gypsum (calcium sulfate 
dihydrate) is used as a soil conditioner, in plaster, cement and plaster 
board, and to make plaster of Paris, the hemihydrate: 

2CaSO 4 2H 2 O = (CaSO 4 ) 2 H 2 O + 3// 2 O 

Calcium sulfide is made by heating gypsum with coke. It is reversibly 
oxidized by air and is not too soluble in water even though it is appreciably 
hydrolyzed to H 2 S + Ca(HS) 2 . A mixture of "lime-sulfur" and water, if 
boiled, produces a mixture of calcium polysulfides which is used to treat 
over-alkaline soils. Simple lime-sulfur mixtures are used as fungicidal 
dusts on fruit trees and garden crops; lime-copper sulfate-water is 
Bordeaux mixture, used for the same purpose. 

Calcium phosphate, Ca 3 (PO 4 ) 2 , is the main constituent in teeth and bone 
and also occurs in large natural deposits which are mined for fertilizer and 
phosphorus chemical manufacture. By reacting phosphate rock with 
H 2 SO 4 , various slowly soluble fertilizers are obtained, "super phosphate of 
lime," for instance, is approximately Ca(H 2 PO 4 ) 2 -2CaSO 4 -2H 2 O. 

Qualitatively, Ca+ 2 is segregated and identified as the slightly water- 
soluble oxalate which is solublized in HC1 to give a red-orange flame test. 


Reactions (in order of decreasing [Ca+ 2 ]) 


Formula Weight 

CaCr0 4 = Ca^ 2 + CrO 4 2 

7.1 x 10~ 4 


Ca(I0 3 ) 2 6H 2 = Ca+ 2 + 2IO^ + 6H 2 O 

1.73 x 10~ 6 


Ca(OH) 2 = Ca+ 2 + 2OH- 

1.3 x 10- 8 


CaSO 4 -2H 2 O = Ca+ 2 + SO^ 2 + 2H 2 O 

2.4 x 10~ 5 


Ca(HP0 4 ) = Ca+ 2 + HPO 4 2 

2 x 10~ 6 


CaC 4 H 4 6 2H 2 = Ca+ 2 + C 4 H 4 Oe 2 

+ 2H 2 

7.7 X 10~ 7 


CaF 2 = Ca+ 2 + 2F~ 

1.7 x 10- 10 


CaCO 3 (aragonite) = Ca+ 2 + CO 3 2 

6.9 x 10~ 9 


CaC0 3 (calcite) = Ca+ 2 + COi 2 

4.7 x 10- 9 


CaC 2 O 4 -H 2 O = Ca+ 2 + C 2 O 4 2 + H 2 O 

1.3 x 10~ 9 


Ca 3 (P0 4 ) 2 = 3Ca+ 2 + 2PO4- 3 

1.3 x 10~ 32 



Quantitatively it is precipitated in similar fashion, then either ignited 
to the carbonate or dissolved in dilute H 2 SO 4 , and the oxalic acid liberated 
is titrated with standard permanganate (see Oxalate, p. 391). Calcium 
in hard water is titrated with versene (Chapter 4) using murexide indicator, 
and "total hardness" (Ca +2 + Mg f2 ) is titrated using versene and 
eriochrome black-T indicator. 

Group 4 Analysis General Description 

Ions of the first three analytical groups are first precipitated as described 
in the preceding chapters. The centrate is evaporated to dryness, HNO 3 
added, the mixture re-evaporated to destroy NHJ salts,* then diluted and 
buffered, and (NH 4 ) 2 CO 3 is added. When the concentrations are correct, 
Mg+ 2 does not precipitate but BaCO 3 , SrCO 3 , and CaCO 3 do. The 
mixture is centrifuged and group 5 ions (Mg +2 , Na+, K+, and NHJ) 
remain in the centrate to be later analyzed for group 5. 

The white residue of group 4 carbonates is dissolved in HAc. The 
solution is buffered with NH 4 Ac, and K 2 CrO 4 is added. This precipitates 
only Ba +a , as yellow BaCrO 4 . It is removed by centrifugation, dissolved 
in HC1, and barium is reprecipitated as white BaSO 4 , which is insoluble 
in HC1 or NaOH and is confirmatory for Ba+ 2 . The Sr^-Ca+^CrO^ 2 
centrate is made alkaline with NH 4 OH, additional CrO 4 2 added, and then 
ethyl alcohol, C 2 H 5 OH. In the mixed water-alcohol solvent, yellow 
SrCrO 4 is not soluble and precipitates, leaving after centrifuging a centrate 
containing Ca+ 2 as the only metal ion. The strontium chromate is 
dissolved in HC1, and the solution gives a characteristic red flame test 
confirming Sr +2 . The Ca+ 2 centrate is boiled to expel alcohol, then just 
acidified, and (NH 4 ) 2 C 2 O 4 added, giving a white precipitate of calcium 
oxalate that may be dissolved in HC1 and the Ca+ 2 is confirmed in that 
solution by a red-orange flame test. 

The procedure with the sample as described is summarized as a flow 
sheet on page 312. An alternate group 4 procedure based on the use of 
ferrocyanide is also given. 


Test solutions contain 10 mg/ml of the group 4 ions. 

Test 18-1. Flame Tests and the Spectroscope. See special experi- 
ment 5, Chapter 23. 

* NH 4 f regulates both CO^ 2 and OH" and a known amount is added later: 
NH^ + OH- - NH 4 OH 
NH+ + COa" 2 - NH 3 4- 




o 1 



Q - eo , co 

* 8 


I o 

S r ^ 


a, Z 


O 52 o __ 





Test 18-2. Some Group 4 Precipitating Agents. Using 6-10 drops of 
each group 4 test soln., try a few drops of the following in separate tests. Record 
formulas and colors of any ppts. that form, and note which reagents might be 
used in differentiating among the test ions: K 2 CrO 4 , Na 2 HPO 4 + a few drops 
of NH 4 CI and NH 4 OH, NaOH, (NH 4 ) 2 SO 4 , (NH 4 ) 2 C 2 O 4 , NaF, and finally 
NaBF 4 + a few drops of NH 4 OH. 

Test 18-3. The Ca+ 2 -Sr+ 2 Separation via the Solubility of Ca(NO 3 ) 2 . 

Calcium and strontium are difficult cations to separate. Two methods are 
given in the procedure, two in special experiment 11, Chapter 23, and three 
more below. 

All the group 4 nitrates are water-solub. but in methyl alcohol (CH 3 OH) the 
solubility at 25 C in mg of salt per ml of alcohol are: Ca(NO 3 ) 2 520, Sr(NO 3 ) 2 
0.6, Ba(NO n ) 2 0.4. Make a mixt. of 5 drops of Ca+ 2 and 5 drops of Sr^ 2 solns., 
add 10 drops of (NH^CO.,, warm, and centrifuge. Discard the liq., add a few 
drops of dil. HNO 3 to dissolve the residue, and rinse it into a small beaker. 
Evaporate the soln. just to dryness, cool, and add a ml of CH 3 OH. Stir well, 
and rinse into a tube with a few more drops of alcohol. Centrifuge and save 
both soln. and residue. Add another ml of alcohol to the residue, stir well, 
centrifuge, and combine solns. in a beaker. 

Evaporate the alcohol just to dryness. Add a ml of H 2 O, 2 drops of NH 4 OH, 
and 6-8 drops of (NH 4 ) 2 C 2 O 4 . A fine, white ppt. is CaC 2 O 4 . 

To the residue from the alcohol treatment, add a ml of H 2 O, 3 drops of NH 4 OH, 
6-8 drops of (NH 4 ) 2 SO 4 , and warm the tube. A slow forming, white ppt. is 
SrS0 4 . 

The above separation may also be made using C 2 H 5 OH, ethyl alcohol, or 
83% (fuming) HNO 3 instead of CH 3 OH as a solvent. 

Test 18 4. The Ca+ 2 -Sr^ 2 Separation via Ca(S 2 O.^) 2 2 . Make a mixt. 
of Ca l2 -Sr+ 2 as before. Add an equal vol. of satd. (NH 4 ) 2 SO 4 and about 
100 mg of Na 2 S 2 O 3 -5H 2 O. Heat in a bath for a few min., centrifuge, and 
save both residue and soln. 

The soln. contains the Ca complex. Add 10 drops of H 2 O, 6-8 drops of 
(NH 4 ) 2 C 2 O 4 , and warm in a bath. The white ppt. is CaC 2 O 4 . It may be 
dissolved in HC1 and a flame test tried. 

The residue is SrSO 4 , contaminated with a little CaSO 4 , and is not tested 

Test 18-5. The Ca+ 2 -Sr+* Separation Using KJFe(CN) 6 ]. Put 10 
drops of Ca +2 in one tube and 10 of Sr+ 2 in another. Add 10 drops of H 2 O 
and about 50 mg of solid NH 4 C1 to each and stir to dissolve it. Now add 3 
drops of K 4 [Fe(CN) 6 ] to each and note that only Ca+ 2 gives a ppt., K 2 Ca[Fe(CN) 6 ]. 

Repeat the test omitting the NH 4 C1 and put the tubes in the hot-water bath. 
Note the same reaction at a slower rate. 

If the Sr+ 2 concn. is large, it too will give these tests, though more slowly. 
Ba+ 2 and Mg+ 2 in large concn. also give the tests, but diln. of the soln. will 
minimize these interferences. 


Make a Ca+ 2 -Sr+ 2 mixt. and ppt. Ca 42 as above. Remove the liq. and add 
(NH 4 ) 2 SO 4 to ppt. Sr+ 2 . This separation is used in the alternate unknown 

Analysis of a Known Mixture 

The class may be directed to analyze a known mixture containing 
group 4 or a mixture of group 4 and 5 cations before proceeding to the 
unknown. If not, then the unknown is begun according to the procedure 
which follows the group equations. 


A. Carbonate Precipitations and Resolutions 

Let M 4 2 be any group 4 metal ion : 

(1) M+ 2 + CC 2 = MC0 3|v 

(2) MC0 3 + 2HAc = M+ 2 + 2Ac~ + H 2 O + CO 2 

B. Barium 

(3) Ba' 2 + Cr0 4 2 = BaCr0 4y 

(4) 2BaCr0 4 + 2H+ = 2Ba+ 2 + Cr 2 O~* + H 2 O 

(5) Ba 42 + S0 4 2 = BaS0 4H , 

C. Strontium 

(6) Sr+ 2 + Cr0 4 2 = SrCrO 4j , (in C 2 H 5 OH H 2 O) 

(7) 2SrCr0 4 + 2H< = 2Sr 42 + Cr 2 O~ 2 + H 2 O 

D. Calcium 

(8) Ca+ 2 + C 2 4 2 = 


(A) If the sample is the centrate from the separation of group 3, it is 
evaporated to 0.5 ml, centrifuged to remove any precipitate of sulfur, then 
put in a small crucible or beaker and evaporated to dryness. When cool, 
1 ml of concentrated HNO 3 is added and the mixture evaporated in the 
hood and heated dry until no more fuming of NHJ salts is evident. The 
residue is dissolved in a mixture of 4 drops of 6 M HC1 and 16 drops of 
H 2 O and rinsed into a 4-in. tube. The crucible is rerinsed with an 
additional milliliter of H 2 O which is added to the previous solution. 
Centrifuge the mixture if not clear and save the centrate. 


Add 100 mg of solid NH 4 C1, then concentrated NH 4 OH until alkaline 
(use a stirring rod to transfer a drop to a drop of phenolphthalein on a 
spot plate), and then add a drop more. Add 12 drops of (NH 4 ) 2 CO 3 and 
put the tube in a nww-water* bath. Stir a few times over the next 
3-5 minutes, then centrifuge, saving the centrate for group 5 analysis and 
the residue for group 4, paragraph (C). 

(B) If the sample never contained ions of the first three groups and includes 
only group 4 or groups 4 and 5 start with 1 ml of it. If it contains NH^~ 
(test 19-8) evaporate with HNO 3 as above. If NH| is not present, pro- 
ceed. Add 1 ml of H 2 O, 200 mg of NH 4 C1, and 2 drops of concentrated 
NH 4 OH. Then add 12 drops of (NH 4 ) 2 CO 3 and place the tube in a 
vvtfAvw-water* bath. Stir a few times over the next 3-5 minutes, then 
centrifuge. If the sample is to be analyzed for group 5 ions, save the 
centrate for that purpose, otherwise discard it. Save the residue for 
group 4 analysis as continued in paragraph (C). 

(C) Stir the residue from (A) or (B) with a mixture of 8 drops of H 2 O and 
2 drops of (NH 4 ) 2 CO 3 , centrifuge, and discard the wash liquid. The 
residue is any combination of BaCO 3 , SrCO 3 , and/or CaCO 3 . Add 1 ml 
of H 2 O and 3 drops of 6 M HAc with stirring. If the solid does not 
dissolve completely, add another drop of acid. Add a drop of phenol- 
phthalein and dilute NH 4 OH until a pink color is obtained, then a drop of 
K 2 CrO 4 . No yellow precipitate indicates Ba+ 2 is absent and one proceeds 
to (). If a precipitate forms, add additional drops of chromate to insure 
complete BaCrO 4 precipitation. Centrifuge, save the residue for (Z)), and 
the centrate for (). 

(D) Dissolve the residue with a mixture of 4 drops of dilute HC1 and 
16 drops of H 2 O. Now put in 4 drops of (NH 4 ) 2 SO 4 and heat the tube in a 
water bath for 5 minutes. Centrifuge and wash the white precipitate 
free of orange dichromate color with a mixture of 1 drop of dilute H 2 SO 4 
in 1 ml of H 2 O. The white, fine, crystalline precipitate is BaSO 4 and 
confirms Ba f2 . It is insoluble in hot HC1. 

() The centrate from (C) contains CrO^ 2 and may also have present Sr 1 2 
and Ca+ 2 . Add 5 drops of dilute NH 4 OH, 3 drops of K 2 CrO 4 , and heat 
it in a boiling-water bath. When hot add dropwise, with stirring, 40 drops 
of 95 % ethyl alcohol. Remove the tube, cool it in a beaker of cold water, 
and stir occasionally. A yellow precipitate is probably SrCrO 4 . Centri- 
fuge, save the precipitate for (F), the centrate for (G). If no precipitate 
appears, Sr+ 2 is absent and the solution is carried to (G). 

(F) Dissolve the precipitate from () in a milliliter of H 2 O, warm, and 

* Too much heating decomposes ammonium carbonate rapidly: 
(NH 4 ) 2 CO 3 - 27V//3 + H 2 O -f CO a 


add 10 drops of (NH 4 ) 2 SO 4 . A fine, white precipitate is SrSO 4 . If no 
precipitate forms in this step after having a yellow precipitate in the 
previous step, the yellow precipitate was K 2 CrO 4 and Sr* 2 is absent. 

Alternately, the SrCrO 4 precipitate may be solubilized in HCl and a 
flame test run. A red flame, contaminated with some violet from K+, 
confirms Sr. 

(C) Transfer the solution from () to a 20-ml beaker and under the hood 
evaporate it slowly to about 10 drops with a direct, moving flame, being 
careful to avoid splattering. If the alcohol vapor catches fire, it is 
allowed to simply burn off. The heating is stopped, and when the beaker 
cools its contents are rinsed into a tube with 1.5 ml of H 2 O. Add enough 
6 M HAc to just decolorize the phenolphthalein, then 1 drop in excess. 
Now heat the tube in a boiling-water bath and when hot add 10 drops of 
(NH 4 ) 2 C 2 O 4 . Allow the tube to remain hot for 8-12 minutes. A white 
precipitate is CaC 2 O 4 , and no precipitate indicates the absence of Ca+ 2 . 
The precipitate may be isolated, dissolved in a few drops of 6 M HCl, 
and a flame test run; a red-orange color confirms calcium. 

This is the end of the group 4 procedure. 


An alternate method for the group 4 separations is desirable because of 
the trouble the Ca-Sr separation gives. 

(H) Precipitate the group as directed in (A) or (), isolate BaCrO 4 as in (C), 
and confirm it in (D). Save the Ca +2 -Sr+ 2 centrate for /. 

(/) Add (NH 4 ) 2 CO 3 until precipitation is complete. Centrifuge and 
discard the centrate. Dissolve the residue of CaCO 3 and SrCO 3 in 6 drops 
of 6 M HAc; add 20 drops of H 2 O and 5 drops of 6 M NH 4 OH. Add 
half a spatula full of solid NH 4 CI. Shake the tube to dissolve the solid. 
Add 6 drops of K 4 [Fe(CN) 6 ], shake, and allow the tube to stand a minute. 
Centrifuge. The residue is potassium calcium ferrocyanide (test 18-5). 
Save the centrate for (7). 

(J) Add 10 drops of H 2 O and 6 drops of (NH 4 ) 2 SO 4 and stand the tube 
in a hot- water bath. A white precipitate is SrSO 4 . 


1. List as many similarities, then as many differences, as you can in the 
chemistries of the group 4 metals. What trends does one detect with increasing 
atomic number? Suppose that radium had not been discovered yet. Predict 


some characteristics for the metal and a few of its compounds. (See also problem 
6 and Chapter 3.) 

2. On the facts gleaned from the preliminary experiments and reading alone, 
suggest an original procedure for the analysis of a solution known to contain 
only group 4 soluble salts, and give it in outline or flow-sheet form. 

3. J. P. Slipshod has made a few cuff notes on group 4 shortcuts: 

(a) "To differentiate between samples of Ba and Ca metals, drop both into 
an aqueous brine solution of 1.7 sp. gr. Barium being heavier will be observed 
to slowly sink. 

(b) "Since the K$ P of CaF 2 is the lowest of the series and the K 8P of BaCrO 4 
is less than that of SrCrO 4 , to analyze an unknown add F~ and centrifuge out 
the CaF 2 ; then add CrO 4 ~ 2 and obtain solid BaCrO 4 , leaving Sr+ 2 in the last 
centrate, and the group is separated completely in two steps. 

(c) "To a solution 10~ 3 M in each metal ion, add enough SO^ 2 to keep its 
concentration 10~ 6 M. This precipitates BaSO 4 . Add more SO^ 2 and main- 
tain it at 10" 4 M to precipitate all the Sr+ 2 , leaving Ca+ 2 in solution to identify 
by a flame test. In this, one reagent does everything." 

Where possible, prepare a quantitative critique of these methods. 

4. What single reagent, test, or observation would enable one to distinguish 
between : 

(a) CaCl 2 and SrCl 2 (b) BaCO 3 and BaSO 4 (c) Sr(NO 3 ) 2 and SrCO 3 (d) BaCrO 4 
and BaCr 2 O 7 (e) CaCO 3 and CaC 2 O 4 ? 

5. A solution is 10~* M in NH 4 OH, 1Q- 1 M in NH 4 + and 10' 2 M in Mg+ 2 . 
Will Mg(OH) 2 precipitate? Explain why Mg does not precipitate in group 4 
as the analysis is conducted here. 

6. From the data given in this text for the first 5 alkaline earths, prepare a 
graph of C vs. atomic number and sketch the melting and boiling point curves. 
Extrapolate the trends to number 88, radium, to predict those physical properties 
of the rarest II A metal. (Ans. Values estimated by this method are 960 C and 

7. (a) Why won't BaCrOj precipitate at low /?H? 

(b) Since HAc dissolves carbonates how does it compare in acid strength 
with H 2 CO 3 ? Explain. 

8. Why does not one use CO 2 gas (as H 2 S gas may be used in earlier groups) 
to precipitate the group 4 ions instead of (NH 4 ) 2 CO 3 ? 

9. (a) Dilute HAc will dissolve SrCrO 4 but not BaCrO 4 . Why? 
(b) Would you expect HCN to dissolve CaC 2 O 4 ? Why? 

10. A group 4 unknown gives a white precipitate with ammonium carbonate, 
another sample of the unknown gives no reaction with potassium chromate, and 
a third portion gives a white precipitate with dilute sulfuric acid. What con- 
clusions can one make? If further tests are needed to analyze the sample, 
explain why and what they should be. 

11. Draw a complete flow sheet for the alternate, (ferrocyanide) procedure 
for group 4 analysis and give the reaction equations for each ion in each step. 

12. (Library) Find reference to the use of potassium rhodizonate in group 4 
analysis. Give reactions, interferences, and sensitivity. 


13. Anhydrous calcium chloride shaken with benzene (C 6 H 6 ) containing a 
trace of water dries the organic liquid by forming hydrates. If shaken with 
alcohol containing a little water, however, one gets only a pasty suspension, 
water is not removed, and some of the salt goes into solution. What explanation 
can you offer for the difference ? 

14. How could one prepare: 

(a) BaCO 3 from BaCI 2 (k) BaCr 2 O 7 from BaCrO 4 (c) BaCl 2 from BaSO 4 
(d) Ca(NO 3 ) 2 from calcite (e) SrSO 4 from SrCrO 4 ? Give the equations. 

15. The solubility of Sr(OH) 2 is such to give a solution 0.148 N in OH~, as 
determined by titration with standard acid. Show that this leads to a calculation 
of the K sl >. 

16. One text suggests that the student use saturated CaSO 4 or SrSO 4 as a 
reagent for Ba+ 2 . What would be the result according to pertinent tabular data ? 


1. M. V. Davis and F. H. Heath, J. Chem. Educ., 27, 626 (1950). (BF~ for the 
Ca-Sr mixture) 

2. R. B. Hahn, J. Chem. Educ.. 30, 349 (1953). (HNO 3 for the Ca-Sr mixture) 

3. G. L. Beyer and W. Rieman Anal. Chem., 19, 35 (1947). (Ba-Sr) 

4. S. Kallmann, Anal. Chem., 20, 449 (1948). (Ca, Ba, Sr) 



The members of what is known as cation group 5 in qualitative analysis 
come from the first two periodic table groups as shown in the figure. 
Ammonium ion is the only cation described on these pages that is not a 
metallic ion, but in many of its reactions it acts like one. Lithium is 
discussed in Chapter 20. It and rubidium, 
cesium, and francium would also be mem- 
bers of group 5, but, the last three, and 
particularly short-lived, synthetic, radioactive 
Fr, are not common in occurrence, or of 
much other than theoretical interest. 




Mg is a silvery metal and the most impor- 
tant of the alkaline earths as an element for 
structural uses. It is the third most naturally 
abundant metal of engineering importance 
(preceded only by Al and Fe), and Mg 4 " 2 is 
preceded in quantity present only by Cl~ and 
Na+ in sea water. 

Mg is produced from sea water and from 
its minerals, carnellite, MgCl 2 -KCl 6H 2 O, 
magnesite, MgCO 3 , dolemite, MgCO 3 - CaCO 3 , 
and brucite, MgO H 2 O. Sea water is treated 



















FIG. 19-1. The periodic 

table in the vicinity of the 

group 5 metals. 


by the Dow process with calcined (CaO) and slaked (Ca(OH) 2 ) oyster 
shells to precipitate Mg(OH) 2 , which is filtered off and dissolved in 
HC1. The MgCl 2 solution is concentrated by evaporation to give solid 
MgCl 2 -2H 2 O which is melted and electrolyzed to give the metal and 
chlorine; the latter, a valuable by-product, is used partly to displace Br 2 
from sea water, and partly to make more HC1. The plants in operation 
each use 20,000 gallons of sea water a minute and 2,500 tons of shells 
a day. 

Isolation of Mg from dolomitic limestones is accomplished by the ferro- 
silicon process described in Chapter 13. 

Magnesium metal is used in light alloys with aluminum (the Dow 
metals) and is all-important in aircraft manufacture. It is harder than 
aluminum and its strength/weight ratio is as good as the best Al alloys, 
but it is more chemically active and needs more surface protection. 
Having a specific gravity of only 1.74, Mg is the least dense structural 
metal. Its m.p. is 650 C and its b.p. is 1126 C. It is also used in incen- 
diary bombs, flares, photoflash bulbs, and as a degasifier in the manufac- 
ture of radio tubes and in the refining of certain metals; it removes 
oxygen from Cu and Ni melts, rids Pb of Bi by forming insoluble Mg 3 Bi, 
desulfurizes Ni alloys, and reduces TiCl 4 to Ti, for example. 

The oxidation states are (1) and (II) but only derivatives of the latter 
are stable and important. The metal is a powerful reducing agent: 

+ 2e- = 2.37 

It burns brightly in air forming the oxide, MgO, and nitride, Mg 3 N 2 . 
Addition of water to this mixture yields the slightly soluble hydroxide, 
Mg(OH) 2 and NH 3 . Salts of Mg <2 are little hydrolyzed, so Mg(OH) 2 
must be a strong base but its alkalinity is considerably curtailed by its 
insolubility. It is soluble in NH+, however, a fact used to retain Mg+ 2 
in the group 5 cation solution, whereas group 4 precipitates. 

Mg (OH) 2 + 2NH+ = 2NH 4 OH + Mg+ 2 

Milk of magnesia is a suspension of Mg(OH) 2 in water. The metal also 
reacts with halogens to form halides, and when heated with sulfur it gives 
MgS which readily hydrolyzes in hot water to Mg(OH) 2 + H 2 S. Metallic 
Mg is used in some reactions with organic compounds, the most important 
of which is the formation of alkyl and aryl magnesium halides known as 
Grignard reagents, which undergo many reactions of synthetic and 
analytical importance: 

Mg + RX = RMgX 

Anhydrous magnesium sulfate is not known but several hydrates 
are (heating them gives a basic sulfate) and these are quite soluble; 



MgSO 4 7H 2 O is epsom salts. Many magnesium compounds form 
hydrates and these aquo complexes, with oxygen attached to Mg 12 , are 
quite stable. Mg(ClO 4 ) 2 , magnesium perchlorate, is one of the best 
dehydrating agents known. 

Mg +2 does not give a visible flame test but is readily detected in group 5 
by precipitation as white, crystalline Mg(NH 4 )PO 4 , or by use of an 
organic reagent described later (test 19-2). 


Reactions (in order of decreasing [Mg+ 2 ]) 


Formula Weight 

MgC 2 4 = Mg+ 2 + C 2 (V 

8.6 x ID" 5 


MgF 2 = Mg+ 2 + 2F- 

8 x 10 8 


MgC0 3 3H 2 O = Mg 42 + COjT 2 + 3H 2 O 

~ 1 x 10~ 5 


MgNH 4 PO 4 = Mg* 2 + NH/ + PO 4 3 

2.5 x 10 13 


Mg(OH) 2 = Mg t2 + 2OH~ 

8.9 x 10- 12 


Mg,(As0 4 ) 2 - 3Mg+ 2 + 2As0 4 3 

- io- 3r> 


Quantitatively, Mg is determined by precipitation as MgNH 4 PO 4 , 
followed by ignition to and weighing as Mg 2 P 2 O 7 , magnesium pyrophos- 
phate. In hard water, Mg i2 is analyzed as described in the last paragraph 
on calcium, Chapter 18, p. 311,>and also colorimetrically, using titan 
yellow with which Mg 42 forms a colored lake. 


Sodium is a metal soft enough to be cut with a knife. It has a pinkish 
tinge and because it reacts rapidly with moisture and oxygen is stored 
in airtight containers or under inactive liquids such as kerosene. Its 
sp. gr. is 0.97, its m.p. is 97.5 C, and b.p. is 892 C. The source of Na is 
sodium chloride as obtained from natural brines, sea water, or salt deposits. 
The metal is produced by electrolysis at about 650 C in the Downes cell 
containing a molten NaCl-Na 2 CO :i mixture. C1 2 is the by-product. If 
aqueous NaCl is electrolyzed the result is NaOH + C1 2 . Metallic sodium 
is used (a) in atomic reactors and sodium-filled valves of high temperature 
engines to aid heat transfer (b) in organic chemistry for certain reductions 
and base-catalyzed reactions (c) to make sodium cyanide, and (</) to alloy 
with lead in the manufacture of tetraethyl lead for gasoline: 

4NaPb + 4C 2 H 5 C1 = (C 2 H 6 ) 4 Pb + 4NaCl + 3Pb 
Additional utilization of the metal is (e) as a metal descaler, following 


heat treatment and (/) as a reductant of T1C1 4 and ZrCl 4 to produce those 
metals. U.S. output of Na metal is 145 thousand tons yearly. 

All the alkali metals are strong reducing agents and the only known 
oxidation state is (I) : 

Na = Na+ + e~ E c = 2.71 

Sodium metal will burn in air when heated forming the yellowish peroxide 
Na 2 O 2 . The latter reacts with water yielding sodium hydroxide and 

One interesting reaction of sodium (as well as other alkali metals) is 
with liquid ammonia. Blue solutions are formed which become bronze 
colored with increasing metal concentration and have very high electric 
conduction. The reaction is 

a:NH 3 + Na -^ Na+ (solvated) + er (solvated) 

The negative particle is the electron, showing the ease with which group 
IA metals ionize. The metal also reacts with hydrogen to give sodium 
hydride, NaH, which is used in metal descaling, with sulfur to form the 
sulfide Na 2 S and polysulfides, and with halogens to give halides, NaX. 
Sodium chemicals are produced and used in large tonnages. Sodium 
hydroxide, one of the strongest bases and industrially the most important, 
is made in Castner cells. A concentrated NaCl solution is electrolyzed at 
a mercury cathode with which Na metal amalgamates and then reacts with 
water to make NaOH. The by-product is chlorine. Sodium hydroxide, 
or caustic soda, is used to make soap by reaction with fats and is also 
reacted with C1 2 to make sodium hypochlorite, NaOCl, a bleach. Other 
uses of NaOH are in the manufacture of rayon by the xanthate process 
where cellulose is dissolved in a CS 2 -NaOH mixture, and in petroleum 
and paper processes. Anhydrous sodium sulfate is called salt cake, while 
the decahydrate, Na 2 SO 4 -10H 2 O, is known as Glauber's salt. Both are 
made from NaCl and sulfuric acid. These salts are used in the paper 
industry to make pulp and as a raw material to produce sodium sulfide 
via carbon reduction. Na 2 S is used to make sulfur-containing dyes and 
for removing hair from hides. The intermediate in salt cake manufacture 
is sodium bisulfate, NaHSO 4 , known also as niter cake. It is used as a 
mild acid in dye baths, metallurgical fluxes, and industrial cleaning prepara- 
tions. Sodium bisulfite is made from a suspension of sodium carbonate 
in water by passing SO 2 through it. The dried product has the formula 
Na 2 S 2 O 5 and is called sodium metabisulfite or sodium pyrosulfite. It is 
used as a bleach in textile manufacture, in food preservatives, and in 
tanning as a reducing agent for chromium solutions. Sodium sulfite, 
Na 2 SO 3 , is produced by boiling the bisulfite solution with more Na 2 CO 3 . 


This may be converted to sodium thiosulfate, Na 2 S 2 O 3 , by heating sulfur 
with its concentrated solution. Sodium nitrate, NaNO 3 , is made from 
sodium hydroxide and nitric acid; sodium nitrite, NaNO 2 , is made from 
the reaction mixture, Na 2 CO 3 + NO + O 2 . The by-products are CO 2 
and some NaNO 3 . Sodium silicates of various ratios of Na 2 O/SiO 2 are 
made by fusing Na 2 CO 3 with sand in a furnace. Silicates are used in 
paste-board adhesives, soaps, textile sizing, fireproofmg, strong alkali 
cleaners, and ceramic glazes. Sodium tetraborate (borax), Na 2 B 4 O 7 10H 2 (X 
is a naturally occurring salt used in cleaners, soaps, and in making other 
borates. Reaction with sodium peroxide and hydrogen peroxide, for 
example, gives sodium perborate, NaBO 3 , which is a bleach and a mild 
oxidizing agent used in cleaners and some mouth washes. Sodium amide, 
NaNH 2 , is made from ammonia and sodium metal. It is used to prepare 
some organic compounds because of its strong reducing and base strengths, 
and to make sodium cyanide, NaCN. The last reaction inferred is between 
NaNH 2 and C and the by-product is H 2 . Sodium cyanide is used as a 
complexing agent in gold mining, as a case-hardening agent for steels, 
in organic reactions to introduce the CN group, and in electroplating 
where it complexes metal ions for slow electrolytic release. Sodium 
carbonate, Na 2 CO 3 , also called soda ash, and the decahydrate, sal soda or 
washing soda. Na 2 CO 3 - 10H 2 O, are made from the ammonia-soda or the 
Solvay process, which dates back to 1869. In this important commercial 
operation, ammonia is reacted with carbon dioxide to make (NH 4 ) 2 CO 3 , 
ammonium carbonate. It in turn is reacted with more CO 2 in water 
solution to yield the bicarbonate, NH 4 HCO 3 , which reacts with NaCl to 
give NaHCO 3 + NH 4 C1. The NaHCO 3 is' heated to produce Na 2 CO 3 
+ CO 2 + H 2 O. Sodium bicarbonate is produced separately from a 
saturated solution of Na 2 CO 3 and CO 2 . NaHCO 3 is used in manufac- 
turing baking powders, carbonated drinks, and fire extinguishers. 

In qualitative analysis, one of the most sensitive and reliable tests is the 
yellow color of the Na+ flame. Many reagent grade chemicals contain 
enough Na+ to give this test. Quantitative analysis using wet chemical 
methods is hampered by lack of redox possibilities and by the fact that most 
sodium compounds are fairly soluble. The usual gravimetric precipitate is 
the sodium zinc uranyl acetate hexahydrate, NaZn(UO 2 ) 3 Ac 9 -6H 2 O. See 
test 19-6. 


Potassium is a silvery metal with a bluish tint. It is similar to sodium 
in chemical action and methods of production, though there is little 
demand for the metal as such. The sp. gr. is 0.86, the m.p. is 63.5 C, 
and the b.p. is 779 C. The sources of potassium are natural brines and 


salt deposits. The principal mineral is sylvinite, a mixture of sylvite, 
KC1, and halite, NaCl. 
Potassium is a very strong reducing agent: 

K = K+ + er E = 2.93 

Like sodium, potassium metal burns in air to form a mixture of oxide, 
K 2 O, and peroxide, K 2 O 2 ; reacts with hot H 2 to form the salt-like hydride, 
KH ; reacts with liquid NH 3 to give blue solutions containing ammoniated 
potassium ions and electrons; reacts with sulfur to give sulfides, with 
halogens to give halides and with H 2 O to give H 2 and the strong base KOH. 

Potassium chloride is the potassium compound in greatest demand 
primarily as a fertilizer component* and as a cheap starting material for 
the preparation of other K compounds. Aqueous KC1 is electrolized to 
make KOH + C1 2 ; the KOH is used chiefly to make potassium (soft) 
soaps and other chemicals. Potassium nitrate for use in black powder, 
fuses, fertilizers, and glass is made by metathesis from NaNO 3 + KC1. 
Potassium acid tartrate, KHC 4 H 4 O 6 , for baking powders is a by-product 
from wine vats. Potassium sulfate for fertilizers is made by the interaction 
of the mineral burkeite, Na 2 CO 3 -2Na 2 SO 4 , with KC1. Potassium 
carbonate, used in making glass, is made by carbonation of a mixture of 
MgCO 3 + KC1 followed by hot-water treatment to decompose the 
product, MgKH(CO 3 ) 2 -4H 2 O, giving MgCO 3 again plus K 2 CO 3 . The 
alloy NaK is a liquid at room temperature in the range 40-90 wt % K. 
It is prepared by reaction between Na and KC1 at 840 C and used as a 
heat transfer liquid and as a reducing agent in some organic reactions. It 
reacts with glass at high temperatures, however, and its reaction with 
polyhalogenated hydrocarbons is explosive. K metal sells for about 
$2.50 per Ib and Na for $0.15 per Ib at the present time. 

A good qualitative test for K+ is the flame test (see special experiment 5). 
There are also a few slightly soluble potassium compounds whose precipita- 
tion is used both qualitatively and quantitatively. These are dipotassium 
sodium hexanitrocobaltate, K 2 Na[Co(NO 2 ) 6 ] ; (NH+ , Rb+, and Cs+ give 
precipitates also with Na 3 [Co(NO 2 ) 6 ]), potassium perchlorate, KC1O 4 , 
and K 2 [SiF 6 ], potassium fluosilicate. 


The ammonium ion, NHJ, is made by reaction of H + with ammonia, 
NH 3 , the combination resulting from dative bonding by the unshared 
electron pair from N. Ammonia is made by the Haber process or its 

* A fertilizer marked for example "10-10-5", contains by weight, "10% N, 10% 
available P 2 O & , and 5% K 2 O soluble in distilled water." according to governmental 
agriculture agencies' definitions. 



higher pressure modification called the Claude process, both of which 
cause direct combination of the elements. The former uses temperatures 
in the range 500-600 and pressures of 100-200 atmospheres over an iron 
oxide catalyst. 

The oxidation state of N in NH 3 and NHJ is ( III). This radical is 
commonly considered in cation analysis because of the frequent occurrence 
of ammonium salts and the similarity of behavior of NHJ and metallic 
ions, particularly K+. The apparent metallic character of NHJ is illus- 
trated in its reduction at low temperatures by sodium amalgam whereby 
free ammonium radicals are formed, and alloy with mercury as would 
a metal. 

NH + <?- = NH 4 

The amalgam explosively releases NH 3 and H 2 on heating. 

Ammonia gas dissolves in water sufficiently to give at laboratory 
conditions a solution containing 28-30% NH 3 or about 60% NH 4 OH 
and having a specific gravity of about 0.90. Ammonium hydroxide is a 
weak base. Ammonium carbonate (NH 4 ) 2 CO 3 ,, is made by reacting CO 2 
gas with aqueous ammonia. Ammonium sulfate, (NH 4 ) 2 SO 4 , for fertilizers 
is made from reaction between aqueous ammonium carbonate and a 
suspension of gypsum. Ammonium dihvdrogen phosphate, NH 4 H 2 PO 4 , 
and diammonium hvdrogen phosphate ; (NH 4 ) 2 HPO 4 , are made by absorp- 
tion of ammonia in phosphoric acid. These salts are used in fireproofing 
wood and textiles and in fertilizers. Ammonium nitrate, NH 4 NO 3 , is 
prepared by reaction between NH 3 and HNO 3 and is utilized in fertilizers 
and in explosives which also contain T.N.T. Ammonium halides are made 
from the halogen acids and NH 3 ; ammonium sulfide, (NH 4 ) 2 S, is made 
from H 2 S and NH 3 and ammonium polysulfides, (NH 4 ) 2 S il ., are prepared by 
dissolving sulfur in (NH 4 ) 2 S solution. 


Reactions (in order of decreasing 
[group 5 ion]) 


Formula Weight 

KC1O 4 = K+ + C1O 4 

8.9 x 10~ 3 


Na 2 [SiF 6 ] = 2Na+ + [SiF 6 ] * 

1.7 x 10~ 4 


NaHCO 3 = Na- + HCO 3 

1.2 x 10- 3 


K 2 [PtCl e ] = 2K+ + [PtCl 6 ]- 2 

1.4 x 10 6 


Na 2 H 2 Sb 2 7 = 2Na+ + H 2 Sb 2 O7 2 

1.3 x 10~ 6 


K 3 [Co(N0 2 ) 6 ] = 3K+ + [Co<N0 2 ) 6 ]~ 3 

10 u 


(NH 1 ) 3 [Co(N0 2 ) 6 ] 

= 3NH 4 ^ + [Co(N0 2 ) 6 ]- 3 

10- 11 

,389. 11 


Analytically, NH is detected by heating with a strong base to drive 
out NH 3 , which is identified by odor, or alternately, K 2 HgI 4 (Nessler's 
reagent), may be added to the mixture to obtain a yellow to brown color. 
Quantitative analysis is made by distilling the NH 3 into a known volume 
of standard acid and back titrating unreacted acid with standard base 
(the Kjeldahl method), or by absorption of the ammonia, in boric acid 
followed by titration of NH 4 BO 2 with standard acid (the Winkler modifica- 
tion). Ammonium salts sometimes interfere with other analyses and are 
removed by heating with HNO 3 or simply by heating; temperatures in 
excess of about 300 C are sufficient to either decompose or vaporize 
(sublime) them. 

Group 5 Analysis General Description 

No systematic separation directions are needed to aid in identifying 
the ions of this cation group since spot tests are available for the detection 
of each member in the presence of the others. 

If only group 5 is to be analyzed in a general mixture, it is first tested for 
NH+ by warming with dilute NaOH and noting the odor of NH 3 . Next 
the first four cation groups are removed using group 3 precipitating 
conditions followed by addition of a mixture of (NH 4 ) 2 SO 4 and (NH 4 ) 2 C 2 O 4 
to complete the separation by removing group 4. The centrate then 
consists of group 5 ions and excess NH^, and is processed as described 

If the sample is known to contain only group 4 and 5 cations, NH is 
again tested for on a separate portion and the main portion is reacted with 
the sulfate-oxalate mixture to precipitate group 4, and the centrate is 
used for group 5. // the sample contains only group 5 cations, NH 4 } is 
tested first again, then spot tests are made for the rest of the group members. 

If NHJ is found, or is introduced in the SO 4 2 -C 2 O 4 2 treatment, it is 
best to remove it by evaporation with HNO 3 , as ammonium ion also gives 
a positive test with one of the few reagents for K + . After baking, the 
residue is dissolved in H 2 O and the solution is ready to be used for tests 
of other ions. See the preliminary tests for these. 


Solutions contain 10 mg/ml of the group 5 ions. 
Test IV 1. Flame Tests and the Spectroscope. 

See special experiment 5. 

Test 19 2. p-NitrobenzeneazoresorcinolforMg* 2 . Put a drop of 
test soln in one spot plate depression, a drop of Sr+ 2 in another, a drop of 




Centrate from 
group 4. 


Traces of 
group 4 


Group 5 

. metals as + Mg l2 \ 

already tested contrlfu e oxalates NHJ 

forNHJ. andsulfates. Na+ | 


Blue lake 

K 2 Na[Co(N0 2 ) 6 ] r 

Mg(NH 4 )P0 4H , 

Flame tests : 
Na+, K+ 

NaMg(U0 2 ) 3 Ac 9 6H 2 0> 

Ba+ 2 in another, and a drop of Mg f 2 in another. Add to each 2 drops of H 2 O, 
1 drop of 6 M NaOH or KOH, and a drop of the org. reagent. Mix each spot 
with a stirring rod, and note and record the differences. See example 15, 
Chapter 12, p. 197. Is this reagent useful as a spot test for Mg+ 2 in a group 4 
and 5 mixt. of ions? 

Test 19-3. Precipitating Agents for Mg+ 2 . Repeat the tests given in 
18-2, and note any similarities and differences between magnesium and the other 
alkaline earth ions. 

Test 19-4. Miscellaneous Group 5 Reagents. With reference to prob- 
lem, 1, this chapter, try one or more reagents made available by the instructor, 
and include data from the literature concerning sensitivity, etc. Note the 
dangers inherent in the use of H 2 SiF B and //C/O 4 ; both cause bad burns, and 
perchlorates are explosive. 

Test 19-5. Blowpipe Test for Mg+ 2 . Using the techniques described 
in part 2 of special experiment 4, and test 17-7, put a spatula full of Mg(OH) 2 
(Mg+ 2 + NH 4 OH) on a charcoal block, add a drop of 0.05 M Co+ 2 soln., and 
direct the oxidizing flame to it with a blowpipe. A pink color possibly due to 


MgCoO 2 develops. Alternately the test may be run using asbestos fiber or 
wadded filter paper in place of charcoal, and heating is done directly in the 
Bunsen flame. 

19-6. Magnesium (or Zinc) Vranyl Acetate jor Na + . Boil 
down 1 ml of Na+ soln. to about one third its original vol., cool, and add an 
equal vol. of Mg(UO 2 ) 3 Ac 8 . What is the slowly forming, yellowish, cryst. 
ppt.? Potassium does not interfere if it is present in quantity less than four 
times that of sodium. The reagent also ppts. ferrocyanide and phosphate, but 
these are presumed absent or removable by methods given in Chapter 2 I . Zinc 
uranyl acetate and some other double acetates work equally well in Na+ pptn., 
but all are appreciably water sol., hence the concn. of the sample. Li + gives a 
similar ppt. 

/Test 19-7. Sodium Cobaltinitrite Jor K+ and 7VHJ-. (a) Solns of 
Na 3 [Co(NO 2 ),j] are not too stable and should be prepd. fresh (0.1 g of reagent 
in 6.5 ml of H 2 O). Shelf reagent should be checked with K + test soln. as follows: 
1 drop of 1 M HNO 3 and 5 drops of reagent are added to 10 drops of soln. 
contg. 0.5-5 mg of K+ and the mixt. allowed to stand 1-5 min to ppt. yellow 
K 2 Na[Co(NO 2 ) 6 ]. The unknown is run in the same way. The reaction with 
NH 4 + is similar and should be tried. Note any difT. in color of ppts. Write 
reaction equations. Give Werner names to all complex compds. 

(b) Preparation of the Reagent (Optional). Dissolve 5 g of NaNO 2 in 6 ml 
of hot H 2 O. Cool to 50 C, add 1.6 g of powd. Co(NO :j ) 2 6H 2 O, and stir. Add 
I ml of glacial HAc, mix, and allow to stand an hour, swirling occasionally to 
aid release of nitrogen oxides. Then add 6-10 ml of alcohol and swirl several 
times during next 10 min. Filter the orange product with suction and wash it 
with 5 ml of alcohol. Spread the filter paper to dry. Bottle the dry product 
and make soln. from it as needed. Reactions are 

Co(NO 3 ) 2 -6H 2 4- 6NaN0 2 = Na 4 [Co(NO 8 )J 4- 6H 2 O + 2NaNO, 
Na 4 [Co(NO 2 ) 6 ] + HNO 2 + HAc = NaAc + H 2 O 4- NO + Na 3 [Co(NO 8 ) e ] 

Test 19-8. Displacement oj NH 9 from NHJ . A strong base will 
liberate a weaker one from its salts. Put 6 drops of NH 4 soln. in a small beaker, 
add 10 drops of H 2 O, and 3 drops of 6 M NaOH. Warm. Remove the flame, 
stir the soln., then cautiously note the odor of ammonia above the container. 
(This test should not be carried out in a tube using an open flame for heat, 
otherwise the soln. may splatter out. Immediately wash off any hot NaOH 
spilled on the skin with plenty of water and very diL HAc.) Further indication of 
NH 3 release may be had by holding a moistened piece of indicator paper above 
the heating soln., and noting evidence of a basic reaction. Rapid heating and 
boiling gives a false indicator test due to tiny droplets of soln. (contg. NaOH) 
being driven out. 

STest 19-9. Nessler's Test for NHj and NH y (a) Dil. a drop of NH 4 f 
test soln. with 2 ml of water, and to a few drops of that, add 2-3 drops of 
Nessler's soln. A yellow to brown color of HO Hg NH Hgl is a positive 


test. The method can be used in quant, colorimetry in the range 0-3 ppm 
NH^ or NH 3 . 

(b) Preparation of the Reagent (Optional). Dissolve 0.03 moles of HgCl 2 
(mol. wt 271.5) in 90ml of hot H 2 O and, with stirring, add to it a hot soln. 
contg. 0.06 moles of KI (mol. wt 166) in 30ml of H 2 O. Filter the red HgI 2 
by suction and wash twice using 30ml of H 2 O each time. Stir the moist 
product into 8 ml of hot H 2 O contg. 0.059 moles of KI. Put the container in a 
water bath and keep it hot for 20 min. Stir occasionally, allowing about a 
fourth of the vol. to evap. Centrifuge or decant the mixt. and discard any residue. 
Put the centrate in an evapg. dish and store over CaCl 2 in a (vacuum) desiccator. 
This will give a moist crystalline mass of potassium tetraiodomcrcuriate(Il) 
dihydrate, K 2 [HgI 4 ]-2H 2 O. Further drying can be done by pressing between 
sheets of filter paper and more desiccation over CaCl 2 . The product is bottled 
and used to make Nessler's solution by dissolving a few crystals in about a ml 
of 3 M KOH. 

Test 19 10. Lithium as a Group 5 Ion. Review lithium chemistry in 
Chapter 20 and optionally perform the tests. Note how one would detect Li+ 
in this cation group if it were included in a general unknown. 

Analysis of a Known Mixture 

Proceed to the unknown sample unless directed to do a known sample 
over group 5. The notebook should be completed to this point. 


A. Magnesium 

1. Mg 42 + NH 4 OH + HPOj 2 = Mg(NH 4 )P0 4|r + H 2 O 

2. Mg+ 2 + 20H- = Mg(OH) 2(| 

3. MgO + CoO = MgCoO 2/ , 

4. Mg 42 + /Miitrobenzeneazoresorcinol = Blue lake (Chapter 12) 

B. Sodium 

1. Na+ + Mg+ 2 + 3UO+ 2 + 9Ac~ + 6H 2 O = NaMg(UO 2 ) 3 Ac 9 .6H 2 O 7 

C. Potassium 

1. 2K+ + Na+ + Co(N0 2 )e 3 = K 2 Na[Co(NO 2 ) 6 ] r 

D. Ammonium 

1. 2NH+ + Na+ + Co(N0 2 V = (NH 4 ) 2 Na[Co(NO 2 ) 6 ] r 

2. Co(NO 2 )e 3 + H 2 O = 2Co+ 2 + 2H+ + 1 1NO 2 + 

2NHJ + 2NO 2 = 2N 2 + 4H 2 O 


3. NH+ + OH- = NH 4 OH = NH 3 + H 2 O 

4. NH 3 + 2HgI 4 ~ 2 + 30H- 


= H-0 Hg N Hg \ Br + 71- + 2H 2 (Nessler's test) 


(A) If the sample is a simple unknown containing only group 5 ions, 
test for NH+ as described in test 19-8. If NH+ is not found, proceed with 
other portions of sample to test described in (). If NH 4 * is detected, 
concentrate 1-1.5 ml of sample to 0.5 ml by boiling in a 20-ml beaker and 
proceed to (C).* 

(B) If the sample is the centrate from previous analysis, NH 4 } is present, 
because NH 4 OH or its salts were added at one time or another and will 
have to be decomposed. There are also possibly some group 4 ions left 
by incomplete precipitation that must be removed so they will not precipi- 
tate as phosphates and render the later Mg+ 2 test inconclusive. These 
interferences are handled in the next paragraphs. 

Boil down the solution available for group 5 in a 20-ml beaker to a 
volume of about 1 ml, add a drop of (NH 4 ) 2 SO 4 and a drop of (NH 4 ) 2 C 2 O 4 , 
and heat the mixture in a water bath for several minutes. Let it cool after 
removal, add 10 drops of H 2 O, stir, then centrifuge any residue and discard 
it unless its volume is appreciable and prior group 4 analysis was not 
satisfactory. (This residue can be solubilized via Na 2 CO 3 transposition, 
as explained in Chapter 21, and returned to the beginning of the group 4 
procedure for another try.) 

(C) To the solution from (A) or (B), add 6 drops of concentrated HNO 3 
and evaporate the mixture carefully in the hood using a moving flame 
under the beaker, and when dry, allow it to cool and add 6 more drops 
of the HNO 3 and evaporate again and heat until no more fumes are 
visible. The NH] salts are now gone.f 

(D) Cool the beaker and its baked contents from (C), add 1 ml of H 2 O, 
and warm briefly to resolubilize the salts that are left. Divide the solution 

* Mg+ 2 may be tested for without elimination of NH/ t and whether NH; 1 is found 
or not, one may perform 19-2 directly. The flame tests for Na+ and K^ may also be 
run at this time although the volatile ammonium salts tend to minimize the K * color 
particularly, and small amounts of K + in the presence of large amounts of NH 4 f could 
be missed, so K^ is always tested for as provided later. 

t No residue indicates no Mg* 2 , K^, or Na< salts were in the sample. Carbon from 
organic matter and silica from the glassware may be visible, but these of course will not 
redissolve in H 2 O. 


equally among four tubes, and proceed as directed in the next four 

() With reference to the four solutions from (D) or (A): 

Tube 1: remove several drops to a spot plate and run test 19-2 for 
Mg+ 2 , if this was not done as mentioned in footnote. 

Tube 2: add 2 drops of concentrated HC1 and run flame tests according 
to special experiment 6, Chapter 23. Use the filter technique if looking 
for K+ in the presence of Na 4 . 

Tube 3: evaporate to about half volume, cool, and run test 19-6 for 
Na + . This precipitation allows one to estimate sodium content. The 
flame test is so sensitive that trace Na f impurities in analytical grade 
chemicals may give a positive reaction. While the flame is visible from 
solutions as dilute as 2 ppm Na+, the solution must contain Na f at a 
concentration of better than 300 ppm to yield the triple acetate precipi- 
tate (without using alcohol to decrease its solubility, which is an un- 
reliable procedure since it may cause the reagent itself to come out of 

Tube 4: evaporate to about half volume, cool, and test for K+ according 
to test 19-7 (a). 

This is the end of the group 5 procedure. 


1. (Library) Some reagents also used in the qualitative analysis of group 5 
ions, but not detailed here, are potassium dihydrogcn pyroantimonate, perchloric 
acid, dipicrylamine, chloroplatinic acid, tartaric acid, hydrofluosilicic acid, and 
titan yellow. Find references to these and list for each the reaction equation, 
evidence that constitutes a positive test, interferences, sensitivity, and any other 
data that appears significant. 

2. An ice plant is using ammonia refrigeration. You are asked to locate a 
pinhole ammonia gas leak in a complex of pipes. What method do you suggest ? 

3. Ammonia leaks are sometimes located by burning a sulfur-containing 
candle, a positive test being the appearance of dense white fumes. Explain the 
basis for this method. Would humidity of the air have any effect ? 

4. If one does not perform the HNO 3 evaporation on the group 4 centrate 
before running the group 5 analysis, what error might be introduced? 

5. One evaporates a 50.0 ml sample of irrigation water to a small volume and 
precipitates K + in it as KC1O 4 , using perchloric acid. Show that, 

ppm K = (mg KC10 4 )(5.63) 

6. Mg is sometimes determined in natural waters as follows: 100.0 ml of H 2 O 


is evaporated to a small volume, Mg +2 is precipitated as (NH^MgPO! and 
ignited to Mg 2 P 2 O 7 which is weighed. Show that 

ppm Mg = (mg Mg 2 P 2 7 )(2.184) 

7. (Library) The Nelson and Castner-Kellner cells are industrially employed 
to make caustic soda. Find reference to these, sketch each, and explain their 
operations briefly. 

8. From the descriptions in this text, give balanced equations for the com- 
mercial production of each of the following: CH 3 MgBr, Na, NaOH, NaH, 
Na 2 O 2 , NaOCl, Na 2 SO 4 - 10H 2 O, Na 2 S, NaHSO 4 , Na 2 SO 3 , NaNO 3 , Na 2 S 2 O : 
Give a use for each one. 

9. As in problem 8: NaNO 2 , Na 2 SiO 3 , NaBO 3 , NaNH 2 , NaCN, 
Na 2 CCV 10H 2 O, (NH 4 ) 2 C0 3 , NaHCO 3 , KOH, K 2 SO 4 , K 2 CO 3 , and KNO 3 . 

10. A colorless, aqueous solution gives no reaction with (NH 4 ) 2 CO :5 + NH 3 , 
and in a separate test gives no flame test and no odor when heated with KOH. 
What conclusions can one make? 

11. What simple tests can one apply to differentiate between: (a) KC1 and 
NH 4 C1 ( Mg(N0 3 ) 2 and NaNO, (r) (NH 4 ) 2 NaCo(NO 2 ) B and K 2 NaCo(NO 2 ) 
(</) halite and sylvinite (e) Mg(NH 4 )PO 4 and MgHPO 4 . 

12. (a) Calculate the pounds of N, P, and K in 100 pounds of a "10-10-5" 
fertilizer. Express each as a per cent. 

(/>) A sample of liquid fertilizer (an aqueous solution of chemicals) has a sp. 
gr. of 1.2. It sells for $2.00/gal and is a 10-10-5 preparation using 80/ by wt 
H 3 PO 4 ($10.00/100 IDS) as a P source, KC1 ($4.00/100 Ibs) as a K source, and 
urea, (H 2 N) 2 CO, ($10.00/100 Ibs) as an N source. If the gallon bottle, label, 
labor, and other fixed expenses are $0.27/gal, calculate the profit realized in 
preparing 1000 gal of solution packaged in gallon bottles. 

1 3. A group 5 solution gives a white gelatinous precipitate and (simultaneously) 
a sharp odor when NaOH is added. What do these observations indicate? 
What confirmatory tests for ions suspected to be present should be run on 
separate portions of sample? On what ions in group 5 have no data been 
gathered by this single test? How should one test for them? 

14. Table salt "cakes" on rainy days due to the presence of traces of hygro- 
scopic MgCl 2 . How could one test such a sample for Mg+ 2 ? 

15. Could any of the group 4 or 5 ions listed for analysis be oxidized to a 
higher state in acid solution by MnOi? Could any be reduced to a lower state 
in acid solution by Sn 11 ? Explain briefly any generalizations you can make 

16. (a) Why is it convenient to put Mg+ 2 in group 5 rather than let it pre- 
cipitate in group 4? 

(b) What is the lowest pH at which it is theoretically possible to precipitate 
Mg(OH) 2 from 1 M Mg+ 2 ? 

(c) Why doesn't Mg precipitate in one of the first three analytical groups? 
(</) Methyl bromide CH 3 Br, reacts with Mg metal in anhydrous ether to give 

the Grignard reagent methyl magnesium bromide. This solution is poured 
over dry ice (CO 2 ) and the reagent opens one carbon oxygen bond and adds 


across it as a positive magnesium bromide ion and a negative methyl ion. 
When water is mixed with that product, magnesium hydroxy bromide and 
acetic acid are formed. Write balanced equations for this illustration of a 
Grignard synthesis. 

17. The solubility of K 2 [SiF 6 ] is about 6 x 10~ 3 moles/liter. Place it in 
Table 19-2. 


1. H. Frmgs, J. Chem. Educ., 21, 460 (1944). (Alizarin for NH^) 

2. W. H. Schecter and J. Kleinberg, /. Chem. Educ. t 24, 302 (1947). (Oxides, 
peroxides, superoxides) 

3. K. H. Gayer and M. J. Elkind, /. Chem. Educ.< 30, 90 (1953). (Inorganic 

4. P. S. Baker, G. F. Wells and W. R. Rathkamp, J. Chem. Ednc., 31, 515 (1954). 
(Alkali metal cell) 

5. M. F. Adams and J. L. St. John, Anal. Chem., 17, 435 ( 1 945). (lodoplatinate for K+) 

6. R. Faber and T. P. Dirkse, Anal. Chem., 25, 808 (1953). (Dipicrylamine for K+) 

7. Inorganic Syntheses Series, McGraw-Hill, New York. 



This chapter outlines the chemistry of eight metals that bear similarities 
to some already described, but these metals are generally unfamiliar to 
the beginner. This does not mean that they are necessarily more rare in 
natural occurrence. Zr, Ti, V, W, and Li are all more abundant in the 
earth's crust than Sn, Zn, Co, Pb, As, Sb, B, Cd, Hg, and Bi, but their 
metallurgy makes them difficult to obtain in a pure state. In recent years 
due to demands in the fields of high-temperature engines, high-speed 
aircraft, electronics, and atomic energy, a tremendous impetus has been 
given metallurgical research, and metals which were laboratory curiosities 
a few years ago are today produced and used in large tonnages. For 
example, zirconium, the twelfth most abundant metal in the earth, sold 
for $300 a pound in 1945 and a grand total of 20 pounds were produced 
in the United States that year. In 1954, because of its desirability as a 
structural material in atomic reactors, about 200,000 pounds of zirconium 
were produced, the price dropped to about $14 a pound, and its metallurgy 
is now better known than that of many metals that have been used 
for years. 

Chemically speaking, the selection of elements in this chapter is not 
particularly representative. Lithium is similar to sodium and magnesium, 
already discussed; beryllium is similar to magnesium and aluminum; 
molybdenum and tungsten are related in properties to chromium ; titanium 
and zirconium have like reactions and oxidation states, etc. 

It is felt, however, that because of their engineering and metallurgical 




importance these metals warrant sufficient emphasis along applied lines 
for inclusion here. 













































FIG. 20-1. The periodic table in the vicinity of the "less familiar metals.' 


Tungsten is a gray, hard metal that is slowly attacked by acids or bases. 
The main ores are wolframite, (Fe, Mn)WO 4 , and scheelite, CaWO 4 . 
These are refined chemically in several ways; the Shoppler process, for 
example, employs fusion with Na 2 CO 3 , leaching with water, filtration, and 
acidification to produce the trioxide, WO 3 . This is reducible with C or H 2 
to give the metal at about 1250 C. H 2 is the preferred reductant since the 
objectionable formation of brittle tungsten carbide cannot be controlled 
with carbon reduction. The method produces tungsten powder, which is 
then compacted and electrically fused. The high melting point of the 
metal precludes casting. The workability of the fused stock is good and 
it may be rolled and formed in the usual ways. W sells for about $4.50 
a pound now. 

The metal is used in alloy steels as a hardening agent and a developer of 


fine-grain structure leading to a densely packed material. Tungsten 
carbides, WC and W 2 C, are among the hardest known alloys and are 
used in cutting tools. Alloys with Fe, Co, Cr, C, V, and Si are common, 
all of them being adapted to service where hardness and high melting 
point are of prime importance: chisels, punches, dies for hot work, lamp 
filaments, etc. Typical alloys are carbaloy, nominally 90% WC, 10% Co, 
and stellite no. 6, nominally 6% W, 39% Cr, and 55% Co. 

Tungsten has oxidation states (0), (II), (III), (IV), (V), and (VI). The 
main oxide therefore is WO 3 . WO 3 is not appreciably soluble in acids 
but will dissolve in bases to form tungstates like Na 2 (WO 4 ), sodium 
orthotungstate, as well as tungstates of more complex structure. Acidifica- 
tion of these hot solutions yields yellow tungstic acid, H 2 WO 4 , and of the 
cold solutions yields a white hydrate, H 2 WO 4 -H 2 O, which changes slowly 
to the other form. The tendency for tungsten-oxygen radicals to con- 
dense is so great that there are very few simple ions of this type in solution. 
Tungsten(III) is known only as chloro complexes in solution, derivatives 
of the acid H 3 W 2 C1 9 . Depending upon concentration and /?H, literally 
hundreds of complex tungsten acids may form and are obviously extremely 
difficult to isolate and study. Tungsten(IV) is known in solution in 
chloro and octacyano complexes. Tungsten(V) is stable in solution only 
as complexes such as W(CN)~ 3 . 

Orthophosphoric acid with normal (WO 4 2 ) tungstates gives heteropolv 
tungstophosphates such as H 3 P( W 3 O 10 ) 4 zH 2 O, where the W 3 O 10 radical 
has substituted for four oxygen atoms in orthophosphate. A wide range 
of radicals* of this type are known, combinations being often represented 
as oxide ratios: 

P 2 5 :W0 3 = 1:24, 1:22, 1:21, 1:20, 1:16, 1:12, 1:7 
As 2 O 5 :WO 3 = 1:16, 1:6, 1:3 
SiO 2 :WO 3 = 1:12, 1:10 

Compounds containing W in the cation, like WF 6 , WC1 6 , and WBr e (the 
only hexabromide known), are not of much importance, since they are 
quickly hydrolyzed in water. In acid solution, active metals like Zn and 
Sn reduce tungstates to blue compounds called "tungsten blues" This is 
a useful color reaction for the qualitative detection of tungsten. (Molyb- 
denum gives a similar reaction.) The tungsten blue structures vary with 
preparation ; formulas as (WO 2 ) 2 WO 4 #H 2 O and W 8 O 23 #H 2 O have been 
suggested. Reduction of tungstates by hydrogen gives a series of tungsten 

* In one nomenclature system, numbers follow names to help differentiation among 
the many compounds. (NH 4 ) 3 [P(W 12 O 40 )] is called ammonium tungstophosphate 


bronzes" whose approximate formula is WO 2 -,rWO 3 , in which jr varies 
from 1 to about 7. These materials are highly colored, some being 
bronze, and all are moderately good electric conductors but are chemically 

Some analysts put W in a special cation group with SiO 2 and a few other 
substances that precipitate in advance of cation group 1 when the sample 
is acidified. If carried along without prior separation it will appear in 
group 1 and may be separated from the other members of the group by 
treating the mixture of chlorides and hydrolysis products with alkaline 
sulfide, in which tungsten alone forms a soluble complex, WS^ 2 . Acidifica- 
tion of the centrate produces brown WS 3 , which hydrolyzes in hot water 
to H 2 W0 4 . 

W is quantitatively determined in steels by dissolving the sample in hot 
aqua regia which converts W to H 2 WO 4 -H 2 O. This is filtered off and 
ignited below 750 C (WO 3 is volatile above that) and weighed as the oxide. 

Molybdenum , 

This element is a silvery metal which can be ductile and malleable 
depending upon temperature and microstructure. Like Ti, Fe, W, etc., 
Mo undergoes an allotropic transformation, but with molybdenum the 
change takes place near room temperature. The main ores are molyb- 
denite, MoS 2 , and wulfenite, PbMoO 4 . These are converted at high 
temperature to MoO 3 , from which the metal is obtained by aluminothermy 
or electric furnace reduction with C, Si, or H 2 . Powdered Mo so obtained 
is compacted and sintered electrically, then, using a sintered bar as one 
electrode and a cooled copper crucible for the other (Mo does not alloy 
with Cu), the Mo is resistance melted at about 2700 C under vacuum or in 
a rare gas atmosphere. Air reacts with Mo at elevated temperatures to 
give both oxides and nitrides and must be excluded during melting. The 
process is typical metallurgy for a number of high-melting metals. Condi- 
tions like 1 100 amperes at 20 voits melting about 30 pounds of Mo an 
hour are common. 

Electrolytic molybdenum of greater than 99.9% purity has recently 
been reported produced on a laboratory scale by the National Bureau of 
Standards. K 2 MoCl 6 and mixtures of alakli metal halides are fused and 
electrolysis is carried out at about 600 C under argon. 

The metal is used mostly as an alloying element in steels to which it 
imparts good air hardening properties that are important in keeping large 
castings from cracking. With nickel, alloys containing up to 30% Mo 
have been prepared for use where corrosion is apt to be severe, as in 
acid-containing reactors. Molybdenum is also used, up to about 4%, 
as a less critical substitute for tungsten in high-speed steels. Such steels 


carry descriptive names like mo-cut, mo-tung, mo-van, etc. Molybdenum 
has been employed in heating elements for furnaces but cannot be used at 
high temperatures without suitable ceramic coatings, such as BeO, which 
protect the metal from forming MoO 3 with the air. Molybdic oxide sub- 
limes and does not give a protective coating to the metal. The largest 
use of molybdic oxide, MoO 3 , is as a catalyst in "reforming" gasoline 
fractions to increase the octane rating by producing branch chain hydro- 
carbons from straight chain ones. An intercrystalline mixture of lead 
chromate and lead molybdate is called "molybdenum orange," one of the 
few important inorganic pigments for paints and inks. Sodium molybdate, 
Na 2 MoO 4 , is used in the order of a few ounces per acre of soil to treat for 
molybdenum deficiency which manifests itself in plants by hindering their 
utilization of nitrogen. Molybdenum disulfide, MoS 2 , is an unusually 
good dry lubricant in that it can be used at temperatures as high as 
1000 C in a nonoxidizing atmosphere as well as at very low pressures. 
Its lubricating ability is not affected by high pressure. Molybdenum 
metal is worth about $2 a pound today. 

Molybdenum has oxidation numbers (0), (IT), (Til), (IV), (V), and (VI). 
The metal is soluble in acids but gives, with HNO 3 , H 2 MoO 4 from which 
white MoO 3 precipitates. This oxide is soluble in bases, forming molyb- 
dates of various formulas like Na 2 MoO 4 and (NH 4 ) 6 Mo 7 O 24 -4H 2 O. 
Molybdic acid, H 2 MoO 4 -H 2 O, like tungstic acid, is capable of many con- 
densation reactions* from which such ions as Mo 3 O^ 4 , HMo 6 O 21 5 , 
HgMo^O^ 3 , etc., have been reported. 

Mo 11 is known in molybdenum "dichloride" [Mo 6 Cl 8 ]Cl 4 , prepared by 
chlorine and the metal. Molybdenum(III) chloride and bromide have 
been made. Heat causes MoQ 3 to disproportionate into the (II) and (IV) 
chlorides. The (TIT) halides hydrolyze in water to form possibly MoO. 
The bonding is largely covalent and the compounds nonionic. Molyb- 
denum(lV) compounds are not stable in aqueous solution or to heat, 
disproportionating into Mo 1TI and Mo v . MoO 2 is made by H 2 reduction 
of MoO 3 . Molybdenum(V) salts are little known, but its complexes are 
stable in water. MoCl 5 is the only pentahalide. It is formed by direct 
chlorination of the metal. H 2 reduces it to Mo + MoCl 3 . Unlike 
tungsten, Mo 2 S 5 and Mo 2 O 6 are known. Molybdenum(VI) is similar to 
chromium(VI) but has a far greater power to form complicated ions, 
since chromium does not condense much farther than Cr 4 O^ 2 . The 
simple molybdates, MoO 4 2 , are known to condense into radicals con- 
taining up to 16 MoO 3 groups in the anion. 

* A condensation is a reaction in which two or more molecules or ions give a more 
complex substance with the elimination of water. A polymerization is the same except 
that water is not a by-product. 


Brown MoS :j is partially precipitated under cation group 2 precipitation 
conditions, leaving a blue color in the supernatant liquid which helps 
identify the element. This "molybdenum blue" is still of doubtful struc- 
ture, apparently being a colloidal dispersion of a molybdenyl molybdate 
hydrate whose approximate formula is MoO 2 (MoO a ) 4 -x'H 2 O. The 
suljide is soluble in alkaline sulfides giving reddish polysuljides like 
MoS 4 2 . 

With a mixture of H 2 SO 4 , excess SCN~, and Sn 11 , Mo vl gives a reddish 
thiocyanate complex of variable composition, possibly Mo(SCN) 5 , which 
is good qualitative evidence for the presence of the metal in solution. 
The thiocyanate ion stabilizes the (V) oxidation state. A reversal of the 
test later described for phosphate may also be used to detect Mo; a 
yellow ammonium molybdo phosphate. (NH 4 ) 3 P(Mo 3 O 10 ) 4 -2HNO a 2H 2 O, 
is formed when MoO 4 3 reacts with PO^ 3 in hot HNO 3 . A number of 
organic reagents are useful in molybdenum analysis; one of the most 
reliable of which is phenylhydrazine in acetic acid. Molybdates give a 
pink to red color. See problem 1 , p. 357. 


This element is a silver/, radioactive metal having three allotropic forms. 
The properties of the allotrones vary considerably. The low-temperature 
form is fairly ductile but presents a unique problem in shaping in that 
heating causes its lattice to expand in two dimensions and contract in the 
third. At 660 C an allotrooic change occurs to a complex, brittle lattice 
which in turn at 770 C gives way to a bodv-centered cubic form that is as 
soft as room-temperature lead. All the forms tarnish rapidly in air and 
become covered with a black oxide, UO 2 . 

The important minerals are pitchblende. U.,OH. and carnotite, a complex 
oxide. Either of these is reducible with magnesium or carbon to give the 
metal. Acid leaching of ores leads to uranium salts bv evaporation of the 
liquors. All uranium compounds are radioactive and poisonous. The 
metal is used almost entirely in the atomic energy program. The 235 
isotope is fissioned directly by slow neutrons, while the 238, the much 
more abundant isotope, is converted via neutron capture and nuclear 
reorganization to neptunium and plutonium. (Pu 239 is neutron fissionable. ) 
U metal sells for about $18 a pound. 

Uranium is chemically quite active. With H at 225 C it forms a 
hydride, UH 3 . Since H is a good neutron velocity moderator, the critical 
mass of the hydride is smaller than that of the free metal itself. The 
metal dissolves in acids; HCl gives H ? and green U IV ; HNO 3 gives some 
H 2 , NO 3 " reduction products, and yellow UO 2 . 

In being oxidized, U->U' 3 , uranium is almost as good a reducing 


agent as metallic aluminum. Oxidation states of (IV) and (VI) are uran- 
ium's most important, but U TTI (purple and green) and (V) (green) are 
also known under special conditions. The (V) salts disproportionate in 
water yielding the (IV) and (VI) states. The most common ion in acidic 
or neutral solution is the yellow, fluorescent uranyl ion, UO;j 2 , and in 
basic solution the diuranate ion, UgO^ 2 , and uranate ion, UO 4 2 . U VI 
forms three kinds of compounds; first, simple ones like UF 6 , second, 
salts derived from UO 3 : the uranates UO 4 2 and diuranates U 2 O 7 2 (which 
resemble chromium and lack the condensing ability of WOj 2 and MoOj 2 ), 
and third, uranyl compounds containing the ion O=U^=O+ 2 . It is 
covalent and characteristic of uranium and most of the elements following 
it in the periodic table. (Np, Pu, and Am have the same four oxidation 
states as well.) Among typical compounds in these series are Na 2 U 2 O 7 , 
which is insoluble in water (as are many U VI compounds) and UO 2 (NO 3 ) 2 , 
which is soluble in ether (as are most UOJ 2 salts). Both compounds are 
used in analytical separations from other cations. The hydroxide, 
UO 2 (OH) 2 -#H 2 O, gives water soluble uranyl salts with acids, like 
UO 2 Ac 2 zH 2 0. 

Uranyl compounds tend to form double and triple salts, as 
NaMg(UO 2 ) 3 Ac 9 -6H 2 O, which are sparingly water soluble and used as 
analytical precipitates. Divalent ions of Co, Cu, Fe, Ni, and Zn may be 
used in place of Mg. Uranyl ion hydrolyzes upon dilution to ions 
containing U and O of indefinite proportions. The volatility of uranium 
hexqfluoride, UF 6 , makes possible the separation of U-235 and U-238 by 
diffusion methods at the Oak Ridge plant. 

Many organic reagents have been discovered that aid both qualitative 
and quantitative analysis for uranium. For example, tannic acid, 
8-hydroxyquinoline, and a-nitroso-/?-naphthol all give precipitates with 
U ions. With excess (NH 4 ) 2 CO 3 , U forms a soluble carbonate complex. 
Under the same conditions Fe and Al in cation group 3 form precipitates. 
The centrate containing the soluble uranium is then rendered basic with 
NH 4 OH and insoluble (NH 4 ) 2 U 2 O 7 falls out. Alternatively, the solution 
is made acidic with HNO 3 , and UO 2 is shown to be present by using 
K 4 [Fe(CN) 6 ] to precipitate brown uranyl f err ocyanide. Several peroxide- 
U systems have been studied a great deal, not only because they are the 
basis for colorimetric methods but also because U ions in aqueous solu- 
tions containing H 2 , O 2 , and H 2 O 2 as products of fission radiation are 
common in atomic plants. 


Titanium is a silver-grey metal with low malleability and ductility except 
at higher than room temperatures. It has two allotropic forms, a, 


hexagonal below 885 C, and /?, body-centered cubic above that tempera- 
ture. The most important minerals are ilmenite, FeTiO 3 , and rutile, 
TiO 2 . The metal is produced in sponge form 99.5% pure by the Kroll 
process in which Mg is used to reduce TiCl 4 at high temperature and 
MgCl 2 is distilled off under reduced pressure. Ti technology is a continual 
source of speculation and invention, and several new methods are 
announced yearly for metal recovery. A recent one involves the reduction 

TiC + 2ZnO 4> CO + 2Zn + TiO 

followed by electrolysis in molten TiO-CaCl 2 . Another method for 
commercial metal production centers about electrolysis of K 2 [TiF 6 ] in 
molten NaCl. Melting the sponge by induction or arc heating in an inert 
atmosphere has permitted preparation of ingots weighing over two tons. 

A preparation of TiCl 4 from the cheaper ilminite ores has recently been 
announced and hinges upon the stability of a titanium Werner ion. The 
ore is dissolved in H 2 SO 4 and iron salts eliminated by fractional crystalliza- 
tion. The TiOSO 4 solution is gassed with HCl at C, then saturated 
with KCI, whereupon K 2 TiCl 6 precipitates. This is filtered off, heated at 
about 400 C, and pure TiCl 4 is evolved. Impurities and KCI are left 

Among the important structural metals, only Al, Fe, and Mg are 
more abundant in the earth's crust than is Ti, but the recovery of Ti is 
hindered greatly by its high melting point (1727 C) and the fact that at 
high temperatures it readily combines with O 2 , N 2 , and many crucible 
materials. Despite these difficulties, its low density (4.5 compared with 
7.86 g/cc for Fe), good corrosion resistance (particularly against HNO 3 ), 
high melting point, one of the highest known metallic strength-to-weight 
ratios, and ability to form many alloys of desirable properties, make Ti 
attractive in engineering uses, and it is rapidly becoming one of the most 
important "new" metals. 

Alloys are made by sintering other metal powders with titanium hydride. 
Hydrogen is released at elevated temperatures to produce a reducing 
atmosphere in the furnace which is desirable, unless hydrogen embrittle- 
ment of metals takes place. Titanium powder is as dangerous to handle as 
magnesium powder, however, since a spark is capable of igniting a fire 
that is next to impossible to put out: Ti + O 2 = TiO 2 . 

Titanium does not absorb lubricants well and is difficult to use in low- 
friction applications. Alloys may be made with good structural properties, 
however, since Ti like Fe undergoes an allotropic transformation which 
allows great alteration of character by heat treatment. The better alloys 
are two phase at room temperature. Titanium is only difficultly surface 
hardened. The best method so far is a nitriding treatment, since the 


solubility of C in Ti is less than 0.1 %. Combinations of TiC, Ti, and 
other refractory materials are sintered for high-ternperature applications 
such as parts in jet aircraft after-burners. Combinations of meials and 
ceramics are given the general name of "cermets." Such materials are 
difficult to form and have low impact resistance but possess the advantages 
of hardness and high melting point. Titanium dioxide is trimorphic like 
SiO 2 and is a dense, brilliant white pigment widely used in paints. TiCl 4 , 
another useful chemical, is employed in smoke screens and sky writing. 

Titanium metal dissolves slowly in HC1 or H 2 SO 4 to yield H 2 and 
titanous oxychlorides or sulfates which have variable formulas due to 
hydrolysis. These compounds are usually repiesented as TiOCl or 
(TiO) 2 SO 4 , but no specific ions like TiO 1 or TiCV 2 have been proved 
capable of individual existence. The metal dissolves in hot HNO :} to 
give titanic acid as an oxidation product, the formula is again uncertain 
but is written H 2 TiO 3 -^H 2 O. TiO 2 is the only Ti oxide stable in water, 
though both TiO and Ti 2 O 3 are also known. Titanium dioxide dissolves 
only slightly in hot concentrated H 2 SO 4 , but when the solid residue from 
that treatment is mixed with water, solution takes place. The solid is a 
titanic oxysulfate. This and all other titanic salts hydrolyze readily so 
that no simple Ti iv compound like Ti(SO 4 ) 2 is ever found in aqueous 
solution. Due to its very weak basicity, titanium forms few normal 
salts of oxygen-containing acids. TiCl 4 can be stabilized only in the 
anhydrous condition or in strong HC1; in the latter the substance present 
is H 2 [TiCl 6 ]. 

The oxidation states of Ti are (II), (ill), and (IV). Little is known 
about Ti 11 except that it is not formed on reduction of Ti iv as is Ti lIT , 
but must be prepared by heating a Ti 111 compound to effect a dispropor- 
tionation to Ti 11 and Ti lv . Ti lv is reducible by reductants like zinc in 
acid solution, giving Ti IIT , itself a good reducing agent as shown, for 
example, by its reaction with sulfite to produce sulfur. Ti 111 resembles 
Cr lff to some extent; all compounds are colored, usually green or violet. 

In acid solution, Ti lv gives peroxy compounds like H 4 TiO 5 (?) with 
hydrogen peroxide. Their yellow to brown color is a good qualitative 
test and adaptable to colorimetry. This behavior resembles that of Zr lv 
and Cr vl with H 2 O 2 . Thiocyanate, acid and Ti TV yields yellow titanyl 
thiocyanates nominally, (TiO)SCN('?). Both color tests are low in sensi- 
tivity if fluoride is present since stable, colorless TiF^r 2 forms preferentially. 
Fluoride added to either of the color tests bleaches the color in proportion 
to [F~]. There are several organic reagents recommended for Ti lv 
analysis. Titanium is analyzed in cation group 3; its hydrated oxide 
Ti(OH) 4 -#H 2 O is not appreciably amphoteric (meta titanates like 
K 2 TiO 3 -4H 2 O hydrolyze readily) so it separates with Fe(OH) 3 -#H 2 O. 


The Fe m -Ti lv mixture is dissolved in H 2 SO 4 , diluted, and H 2 O 2 added. 
The characteristic peroxy titanium color is observed even in the presence 
of ferric ion. 

Quantitatively, Ti is determined after removal of interfering elements 
like Fe, by colorimetric analysis of the peroxy color, or by titration of 
Ti 111 with standard MnO 4 . 


Zirconium is a silvery metal that is malleable and ductile at room 
temperature when pure and fully annealed. It has an allotropic trans- 
formation at 865 C, at which temperature hexagonal oc-Zr changes to 
body-centered cubic /?-Zr. Zr reacts with air at elevated temperatures 
and burns brightly to the dioxide. There are two important minerals, 
baddeleyite, ZrO 2 , and zircon, ZrSiO 4 . Both contain hafnium as an 
impurity and since this metal is very like Zr (the most similar pair of 
elements in the periodic table), it is carried through metallurgical processes 
and found in all commercial Zr. Zirconium is ihe eleventh most abundant 
element in the earth's crust, its quantity estimated to be more than the 
combined amounts of Cu, Ni, Pb, Hg, and Zn. 

Two methods are used to obtain Zr metal. The older de Boer-Van 
Arkel process (1924) produces a high-grade product by the thermal 
decomposition of ZrI 4 in which the metal collects on electrically heated 
wires of pure Zr. The newer Kroll process (1947) yields slightly less pure 
metal but is cheaper to operate and capable of larger production. This 
process utilizes a Mg reduction of fused ZrCl 4 at high temperature, giving 
a sponge of Zr from which the salt residue is melted away in a vacuum 
furnace. The sponge is compacted and melted by resistance at about 
1875 C in an inert atmosphere. 

Getting rid of hafnium, found in all zirconium ores, is the main benefi- 
ciation problem. The most recently announced process utilizes a separa- 
tion based upon solubility differences of some Werner complexes. The 
mixed Zr and Hf tetrachlorides are dissolved in water giving ZrOCl 2 , 
HfOCl 2 , and HC1. To this is added NH 4 SCN and NH 4 OH, yielding 
thiocyanato complexes of ZrO+ 2 and HfO+*. The hafnium compound is 
soluble in an extracting solvent, methyl isobutyl ketone. The Zr complex 
remaining behind is decomposed with H 2 SO 4 and zirconium hydroxide is 
precipitated using NH 4 OH, and separated by filtration. It is fired at 800 C, 
giving ZrO 2 which is chlorinated to ZrCl 4 , ready for Mg or Na reduction. 
Sodium is replacing magnesium as a reductant today. Zr sells now for 
about $4.50/lb. 

Zirconium is used in a few alloys that are interesting because of the 
moderate density of Zr, the ease with which it alloys, and the high hardness, 


corrosion resistance, and melting point of the product. Some relatively 
high-temperature (200-400 C) Mg and Zn alloys contain zirconium. 
The metal, as such, is used extensively in atomic installations. In addition 
to possessing good structural features, Zr is transparent to thermal 
neutrons and neutron economy is important in nuclear reactors. Zr 
dissolves gases in large volume and is employed for this purpose in vacuum 
tubes. Zirconium boride is a refractory used at temperatures above 
3000 C, the nitride with a melting point of 3200 C is stable and electrically 
conducting. Zirconia, ZrO 2 , is made by heating dolomite and zircon 
sand. It is used as a furnace lining material. 

The metal is not attacked by dilute acids, concentrated HC1, or any 
bases, but will dissolve slowly in hot concentrated HF, H 2 SO 4 , or H 3 PO 4 . 
The metal forms nonadherent ZrO 2 in moist air at elevated temperatures, 
but, at moderate temperatures and below, its resistance to attack is 
comparable to 18-8 stainless steel. 

Zirconium has oxidation states of (II), (III), and (IV), but only the 
last is stable in aqueous solution ; ZrCl 3 put in water, for example, liberates 
H 2 and forms a hydrolyzed Zr lv mixture. The halides (except F) of 
Zr 11 are known and are all very dark colored compounds. When heated 
they disproportionate into Zr + Zr lv . Zr m forms halides also of Cl, Br, 
and I, and these are the only well-defined zirconous compounds prepared. 
Zr lv salts like those of Ti lv are strongly hydrolyzed, and it is not possible 
to crystallize simple salts like Zr(SO 4 ) 2 from water. Hydrolysis gives 
basic salts but as in titanium chemistry, no single ions like ZrO+ or ZrO+ 2 
are observed and definite formulas are not given. For the sake of 
simplicity, however, one may say the ZrO+ 2 , zirconyl ion, will serve to 
describe the behavior of Zr ]v in water solutions. 

Zirconates, ZrO 3 2 , are formed when the oxide ZrO 2 is fused with 
alkalis, much as titanates and silicates are made from TiO 2 and SiO 2 . 
Zirconate salts have low solubility, as contrasted to zirconyl compounds. 
With oxalate, ZrO+ 2 gives white ZrO(C 2 O 4 ) which dissolves in excess 
reagent to form ZrO(C 2 O 4 ) 2 2 . From hot solutions strongly acidified 
with HNO 3 , any phosphate, including H 3 PO 4 , will precipitate zirconium 
as ZrO(H 2 PO 4 ) 2 -H 2 O. This precipitation is almost specific for Zr in a 
general ion mixture. In cation group 3 where Zr is normally analyzed, 
its hydroxide is not particularly amphoteric and is insoluble in dilute 
NaOH, classing it with Fe m and Ti lv . It may be separated from these 
by adding HNO 3 and H 3 PO 4 . In this procedure Fe(OH) 3 and Ti(OH) 4 
dissolve while Zr lv forms the insoluble, gelatinous phosphate. Alter- 
nately Zr lv may be separated from other group 3 cations by means of 
(NH 4 ) 2 S + (NH 4 ) 2 C 4 H 4 O 6 , with which combination Zr lv forms a soluble 
tartrate complex, whereas the other metals form insoluble sulfides and 


hydroxides. From the centrate, Zr lv can be precipitated as the zirconyl 
dihydrogenphosphate hydrate which in quantitative work is filtered off and 
ignited to ZrP 2 O 7 . 

The fluoride test with alizarin, as described under fluoride among the 
anion analyses, may be reversed to test for Zr; a solution of ZrO r2 and 
alizarin red-S in acid solution give a red color which is bleached yellow 
if F~ is added. The fluoride is capable of decomposing the Zr-dye lake 
by forming the more stable ZrFg 2 . Zr compounds do not give colored 
bead tests, but when subjected to the blow pipe-dilute Co ' 2 -charcoal 
block technique explained previously under Al and Zn, a faint yellowish 
green, reminiscent of Rinmann's green, appears. 


Vanadium is a steel grey, slightly magnetic, brittle metal having the 
highest hardness of any element except the diamond allotrope of carbon. 
The principal ore is patronite, the empirical formula for which is approxi- 
mately VS 4 . Vanadium is common in small amounts in many rocks and 
uranium minerals and is recoverable also from concentrates like slags and 
ash heaps. Minor ores include vanadinite, Pb 4 (PbCl)(VO 4 ) 3 , and roscoe- 
lite, H 8 K 2 (Mg, Fe)(Al, V) 4 (SiO 3 ) 2 . Vanadium is obtainable by the 
Goldschmidt process, but since no use is made of the pure metal, its ores 
are frequently mixed with those of iron, reduction of the mass made, and 
the mixture of Fe and V used as an additive in steel manufacture, which is 
its main use. The finished steel may contain up to 4.8 % V and finds wide 
acceptance for cutting tools in alloys like vanite high speed, supervan 
dreadnought, etc. Vanadium has good solubility in many metals and is 
successfully alloyed with Al, Be, B, Nb, Cr, Co, Cu, Fe, Mo, Ni, Sn, Ti, 
W, and Zr. 

The main oxide is V 2 O 5 , which is a contact catalyst in the high tempera- 
ture oxidations of SO 2 to SO 3 and naphthalene to phthalic anhydride. 
V 2 O 5 is a brown to black material while the next lower oxide, V 2 O 4 , is 
blue. The oxide VO 2 is also known, but VO is not. 

The chemistry of vanadium resembles that of phosphorous and it is 

The oxidation states of vanadium are (V), (IV), (III), and (II) in that 
order of importance, although numbers of compounds of each are known. 

V 2 O 6 can be made by heating ammonium metavanadate^ NH 4 VO 3 . It is 
slightly soluble in H 2 O, giving an acidic solution and hydrolyzes to 
produce complex vanadates bearing some resemblances to phosphates and 
some to chromates. The formula (V 2 O 5 ) w -xH 2 O is used to describe the 
condensation products. With H 2 O 2 and acid, peroxyvanadates like VO^ 
form; these solutions are yellow to red in color and are a color test for 


V v . Bases and vanadates yield condensation products as V 2 O~ 4 , 
V 4 0r 3 6 , V^, etc. 

V v is easily reduced to V 1V , an oxidation state which occurs in three 
common ways: vanadium(lV) halides as VC1 4 , vanadyl salts as VOSO 4 , 
and vanadites as K 2 V 4 O 9 . 

V 111 is obtained on reduction of either V v or V 1V . It resembles Fe^ 3 . 
It forms salts and can be present in solution as hydrated, green V i3 
which then hydrolyzes to give oxygenated products, the ion VO 4 
presumably being present in them. 

V H is not produced in the presence of air or water because of its reduc- 
ing ability and not many compounds are known. A typical preparation 
is the heating of VI 3 to cause disproportionate to I 2 and VI 2 with 
reaction carried out in an evacuated tube. The peroxy vanadate color 
already described does not fade in the presence of H 3 PO 4 or F~ as do 
colors of Ti and Fe compounds with the same reagents, and hence serves 
as a qualitative test for V as it appears in cation group 3. In basic solution, 
H 2 S converts VO^ 3 to VS^ 3 , having a characteristic purple color from 
which some brown to black oxv suljides of vanadium like Na 4 V 2 O 2 S 5 
precipitate upon acidification. Apparently the simple V 2 S 5 has not been 
observed. The supernatent liquid from that treatment will appear blue 
due to VO+ 2 formed under the reducing action of H 2 S in acid. These 
color changes qualitatively identify V in group 3. 

It is quantitatively run by reduction in H 2 SO 4 solution to V 1V , using 
SO 2 , followed by expulsion of the gas by CO 2 and titration at 70 C to 
V v with standard permanganate. 


A silvery, brittle metal, Be is obtained from its only important mineral, 
beryl, Be 3 Al 2 (SiO 3 ) 6 . Because beryl has about the same density as the 
gangue, no simple beneficiation methods are known, and the supply of 
beryllium ore for processing is dependent upon hand selection of pieces 
of worth-while size. Beryllium melts at 1350C and its metallurgy is 
beset with difficulties. One method of obtaining the metal is via conver- 
sion of beryl to BeCl 2 or BeF 2 , fusion with NaF, and electrolysis of the 
melt. These halides of Be, however, are very hygroscopic and the 
hydrolysis products like Be(OH)Cl #H 2 O do not give the original materials 
again upon heating. Beryllium metal has a sp. gr. of 1.86 and tends to 
form droplets which resist coalescence and float on the top of the bath 
where they oxidize and occlude impurities. Another method of obtaining 
the metal is the thermal reduction of BeF 2 with Mg, giving a large aggregate 
of Be together with a fusible slag that is melted away under vacuum. This 
is essentially the Kroll method again. The crude lump is vacuum melted 


and purified to metal selling now for about $55 per Ib. Molten Be is 
very reactive and fusions are conducted under special atmospheres in 
BeO crucibles, which themselves are a major production problem. 
Beryllium oxide is a refractory material that is formable by pressure only 
in graphite molds at high temperatures. At present, ingots of Be are 
produced mostly by powder metallurgy and are not easy to work due to 
the brittle character of the metal. 

A more recently reported extractive method is the Sheer-Korman 
process. Raw ore is mixed with one-third its weight of soft coal and 
shaped and baked into electrically conducting rods. An arc is struck 
between them, giving temperatures reported in excess of 8200 C. Elements 
from the ore issue from the arc as a high-speed flame and contact C1 2 in 
the main part of the furnace, producing chlorides. These are fractionally 
condensed in cool zones lo effect separations of various constituents. 
BcCI 2 is then electrolv/ed in a molten bath to produce a nearly pure, 
nonbrittle metal which is machinable and estimated to cost about half 
that of metal produced bv older processes. 

Bervllium has two important uses, one in cooper, nickel, and aluminum 
alloys and one in the atomic energy program. In the alloys. Be imparts 
considerable strength. A 3, Be-copper alloy is age hardenable, six times 
stronger than pure copper, nonmagnetic and nonsparking, and used widely 
in the electrical and machine industries for tools, springs, diaphrams, 
heater controls, connectors, cams, gears, oil seals, gauge elements, etc. 
Bervllium is a bettor electric conductor than copper In the atomic 
program. Be and BeO are used as neutron reflectors, thus enabling a 
neutron-producing mass smaller than normal to be critical, because 
neutrons escaping the mass are reflected into it. This mav be the principle 
by which relativelv small atomic artillerv shells function. Be metal, BeO, 
and the soluble compounds are verv toxic. 

The oxidation state of the element is (II). and it acts in solution as an 
intermediate between aluminum and magnesium. Both the metal and 
oxide are slowlv soluble in acids. 

Beryllium has a marked tendency to form four covalent bonds, and it 
has a stronger ability to hvdrate than any other dipositive ion due to its 
small size. Covalencv explains the phenomena of BeO being insoluble 
in water but soluble in Be 11 salt solutionsthe oxide coordinating with 
the ion giving groups like (Be O->Be) t2 to build up polymeric ions. 
Evaporation of these gives gels of no special stoichiometry. The 
hydrolvsis of Be' 2 in water is virtually complete; therefore, one cannot 
obtain bervllium salts of weak acids from solution. 

BeS is not hydrolvzed in H 2 O as is A1 2 S 3 , but, like AUOH),, Be(OH) 2 
is amphoteric and dissolves in bases to furnish oxygenated ions such as 


BeO(OH)-, and is formed in preference to the sulfide in alkaline sulfide 

Be(OH) 2 normally appears in cation group 3 and is separated from 
A1(OH) 3 , utilizing the fact that Be(OH) 2 is soluble in hot 1.5 M NaHCO 3 
whereas A1(OH) 3 is not. Another method is to precipitate Fe+ 3 and AP 3 
from an acetate buffer at/?H 5.5 with 8-hydroxyquinoline, then to precipi- 
tate Be 11 in the centrate as Be(OH) 2 with NH 4 OH. For quantitative 
determinations, the hydroxide is ignited to BeO, which is weighed. 


Lithium is the lightest metal. Its sp. gr. is 0.53, its m.p. is 186C, 
and b.p. is 1336 C. It is harder than sodium or potassium but softer at 
room temperature than lead. The crystal structure, like that of other 
alkali metals, is body-centered cubic at room temperature, but, aided by 
pressure, at 196C it undergoes an audible lattice change to a face- 
centered structure. Two mixed silicates are mineralogically important, 
lithium mica, K 2 Li 2 Al 2 Si 3 O 9 (F, OH) 2 , and spodumene, LiAl(SiO 3 ) 2 . In 
addition, mixtures of salts containing some Li 3 (PO 4 ) are found in 
saline deposits. The free metal is obtained by electrolysis of its fused 

Per se, lithium is used in a few light alloys as a hardening agent and 
with iron as a mixed catalyst for the ammonia synthesis, but beyond that 
the metal has little application. Lithium compounds on the other hand 
are finding wide use and frequently the demand runs ahead of supply. 
Lithium deuteride, LiD, is probably used in the hydrogen bomb, this 
crystalline solid allowing a relatively large mass of those light elements to 
be packed close to the reaction center to insure fusion. Temperatures 
then reached are guessed as high as 10 8 C. Whereas the critical mass of 
a fissionable element may be small, LiD is handled in inert atmospheres 
in any mass, leading to awesome speculation concerning the power of 
thermonuclear devices. Lithium aluminum hydride, LiAlH 4 , is a powerful 
reducing agent, having a specificity of reaction toward certain reducible 
groups. Lithium greases are used in military vehicles because of their 
good performance under drastic conditions; Li 2 CO 3 is used in glass and 
porcelain manufacture to produce a high index of refraction; LiCl is 
used in drying systems; LiF is used as a welding flux; lithium stearate is 
used in cosmetics and waxes; Li 2 CrO 4 is alcohol-soluble and used as a 
rust inhibitor in radiators, lithium aluminum silicate is a filler in rubber, 
lithium hydroxide monohydrate is the electrolyte in one Edison type battery, 
lithium chlorate is used to impart a red color in pyrotechnics, lithium 
cynide is used in plating, lithium perchlorate is an oxidant in solid rocket 
propellants, etc. 



The chemistry of lithium is something like that of Na h and K 4 but 
also bears resemblance to that of Mg i2 and Ca f2 . Its only valence state 
is (I). Lithium may be melted and poured in air without tarnishing 
whereas the other alkali metals cannot. Molten, however, it shows an 


Half-Cell Reaction 


W + 2H 2 O = W0 2 + 4H+ + 4e~ 


W + 8OH~ = WO 4 2 + 4H 2 O + 6e~ 


Mo + 8OH- = MoO 4 2 + 4H 2 O + 6c~ 


Mo + 4H 2 O = H 2 Mo0 4(aq) + 6H+ + 6e 


(MoO 2 ) 2 MoO 4 + 4H 2 O = 3H 2 MoO 4 + 2H^ + 2e~ 


U = U t3 + 3e~ 


U+ 4 + 2H 2 O = UO 2 4 2 -1- 4H+ + 2e~ 


Ti -f 2H 2 O = TiO 2 (hydrated) + 4H^ + 4e~ 


Ti+ 3 + H 2 = TiO+ 2 -f 2H+ -f e~ 


Zr + 2H 2 O = ZrO 2 + 4H^ + 4e~ 


Zr = Zr+ 4 + 4e~ 


VO+ 2 -f 3H 2 O = V(OH) 4 I + 2H+ -f e~ 

- 1.00 

V+ 3 + H 2 O = VO+ 2 -f 2W 4- e~ 


V+ 2 = V+ 3 + e- 


V = V+ 2 H- 2e~ 

^ 1.18 

Be = Be+ 2 + 2e~ 


2Be H- 6OH- = Be 2 O^ 2 + 3H 2 O + 4e~ 

2.62 (ED 

Li = Li+ -f e~ 


* Reprinted by permission from W. M. Latimer, Oxidation Potentials, 2nd ed., 
Prentice-Hall, New York (1952). 










Per (oxy) tuncstate 

WO 4 
W 7 24 
W 4 13 
WO 2 

- 2 
- 6 

- 2 



T.0 3 





W )Z 41 

- 10 











BeHJll) 4 


Mo 3 n 
Mo 7 O 24 

P(Mo 3 10 ) 4 
Mo0 2 

._ 4 
- 6 

- 3 



'/ r r\ 






z,ru 3 


v w 4 


V0 2 
V 2 7 

v,o lfi 

V0 4 



- 4 
- 7 
- 1 

Uranium! I Vj 

UO 2 
U 2 7 
U0 4 





V0 3 

V 2 7 

- 1 
- 4 




* One of several known variations 

avidity for nitrogen and forms a nitride, Li 3 N, reminiscent of magnesium 
action. Li metal reacts only slowly (like Mg, unlike Na) with water, and 
gives only Li 2 O (no Li 2 O 2 , lithium peroxide) with hot oxygen. By way of 
further contrast, Li f complexes more readily due to its small size and 
forms some hydrates, alcoholates\ amnioniates, and neutral complexes 
with organic reagents; its halicle salts, except fluoride, are more soluble 
and its carbonate, phosphate, and hydroxide are less soluble than those of 
other metals in group 1 A of the periodic table. 

Lithium is analyzed in cation group 5 usually by two methods. The 
first is to boil the sample just to dryness with HCl, then extract LiCl with 
acetone, or alcohol, leaving the other chlorides behind. A flame test on 
the extract will be a brilliant red if lithium is present. The second method 
hinges on precipitation of Li^ as Li 3 PO 4 from an alcohol-water mixture 


followed by a flame test on the washed residue. The precipitating agent 
is NH 4 (H 2 PO 4 ) + NH 4 OH. and while Mg may also precipitate as 
Mg(NH 4 )(PO 4 ) if not removed in group 4, its presence will not alter the 
conclusive Li flame test. Tn quantitative analysis, LiCl is separated by 
organic solvent extraction as above after removal of interferences, then 
heated to dryness with H 2 SO 4 to produce Li 2 SO 4 which is weighed. See 
also test 20-23 for a colorimetric method. 


If samples of the eight metals as well as specimens of other rarer metals 
like tantalum, hafnium, niobium (columbium), etc., are available, examine 
and characterize them briefly in the notebook. Examine a few of their 
salts also, noting color, formula, and other features of each. 

Most of the following laboratory experiments use a few milligrams of 
dry salt, metal, or mineral for each test. The student should attempt to 
write the equations for all reactions described and should make a note of 
the characteristics of each reaction, including speed, colors, types of 
precipitates, etc. 

Test 20 1. Tungsten Blue. Dissolve a crystal of H 2 WO 4 H 2 O in a 

few drops of dil. NH 4 OfT then add JO drops of coned. HO and a few granules 
of Zn or a few drops of 25% SnCl 2 in coned. HC1, and stand the tube in a warm 
water bath. Note the development of color over the next 5- 15 min. (The 
blue color docs not disappear with excess SnCl^ 2 , whereas the analogous 
molybdenum blue does fade.) The concn. limit is about 50 ppm W. 

Test 20-2. W in Steel, (a) Dissolve at least 10 mg of sample in aqua 
regia. Any residue is probably C, SiO 2 and/or WO 3 . Remove the soln., treat 
the residue with a few drops of coned. NH 4 OH, stir, add a ml of H 2 O, and 
centrifuge. Remove the soln. (which contains tungstate), acidify with HC1, 
and run test 20-1, or acidify with coned. H 2 SO 4 and run test 20-3 on the 

(b) Try this as a metal surface spot test if a larger steel sample contg. at least 
1 % W is available: file a spot clean, then etch it with a drop or two of aqua 
regia made from 6 M acids. After a few min., carefully remove the drop with a 
dropper and blot with a piece of filter paper. Add a drop of 6 M HCl to the 
spot and remove it after several min, as before. Put a piece of filter paper over 
the spot, add a drop of H 2 O and a drop of 25% SnCl^ 2 reagent. The color 
intensity is an approx. measure of W content. 

Test 20-3. The Defacqz Method for W. Mix a small amount of a 
W-contg. solid, such as a mineral or a salt, with 5 times its wt of KHSO,, and 
heat gradually to melt and fuse the mass. Keep it molten for 5-10 min. Cool. 
Add several drops of coned. H 2 SO| and mix well. Add a crystal or two of 


hydroquinone C 6 H 4 (OH) 2 . Tungstates give a violet color reaction but molyb- 
dates do not. The concn. limit is 20 ppm W. 

Test 20-4. Tungstic Acid from Wolframite Ore. Fuse 5 parts of 
powd. wolframite with 8.5 parts of anhyd. Na 2 CO 3 and 1.5 parts of NaNO 3 in 
a suitable container like a Pt crucible. After the mass has cooled, extract it 
with hot water and filter. To the filtrate, add excess HC1 to ppt. H 2 WO 4 H 2 O. 

Try the bead test in special experiment 4 for W minerals, using microcosmic 

Test 20-5. Mo in Minerals Using SCN~. Make a test tube from a 
3-in. piece of 8-mm soft glass tubing by melting one end shut and turning a lip 
on the other end with the aid of a file. Make an intimate mixt. of 250 mg of 
Na 2 CO 3 , 250 mg of KNO 3 , and lOOmg of powdered mineral. Pour into the 
tube and heat gradually, finally fusing at a red heat for about 5 min. While 
the tube is still hot, touch the bottom in 8 ml of H 2 O contained in a 50 ml 
beaker to break the tube. Pick out the pieces of glass and discard them in the 
waste jar. Heat the rest of the mixt. and boil gently to evaporate 2-3 ml of 
soln. during the next 5 min, then pour it into two test tubes and cent. Combine 
the centrates in a small beaker, add a drop of phenolpthalein, then add 1 M 
HC1 dropwise until the red color fades. Add 200 mg of Na 2 (C 4 H 4 O 6 ) 2H 2 O, 
which will form a tartrate complex with any tungsten present. Add 12 drops 
of coned. HC1 and shake. Add 6 drops of KSCN and 10 drops of SnCl^ 2 
solution (made by dissolving 10 g of SnCl 2 -2H 2 O in 100 ml of 2 M HC1). After 
1 min, add 6 drops of diisopropyl ether and shake. A blue organic layer 
indicates molybdenum. This method is adaptable to quant, colorimetry in 
field testing, in the range 0-32 ppm molybdenum. See reference 6, p. 356. 

Test 20-6. Molybdenum Blue. Dissolve a few mg of ammonium 
molybdate in a few drops of coned. HC1 and dil. to 2 ml with H 2 O. Put 1 drop 
of this soln. in one test tube, 2 drops in a second, 5 drops in a third, and 10 drops 
in a fourth. To each of these add enough H 2 O to give a total vol. of 2 ml, 
then to each add 3 drops of 0.2 M (NH 4 ) 2 HPO 4 . To each tube add 2 drops of 
0.25 M SnCl^ 2 . The blue colloidal suspension has the approx. formula 
Mo 8 O 23 -#H 2 O and is analogous to tungsten blue. 

Test 20-7. Mo in Steel. Dissolve a small steel shaving in aqua regia, 
dil. to about 2ml, and centrifuge; the residue is probably C, SiO 2 , and/or 
WO 3 . Discard the residue. Add excess 6 M NaOH to the soln. and centrifuge ; 
the residue is primarily Fe(OH) 3 with molybdates retained in soln. Reacidify 
the soln. with 6 M HC1. Use portions of the soln, for the test below and also 
for previously described Mo tests. 

Put a few drops of soln. on a spot plate, add a drop of 6 M H 3 PO 4 , a crystal 

s \ 

of potassium xanthale, C O C 2 H 5 , and a drop of 3 M HC1. A violet color, due 

probably to MoO 2 [SC(SH)OC 2 H 5 ] 2 , indicates Mo. The concn. limit is 3 ppm. 


Test 20-8. V in Steel. Dissolve a small shaving in a mixt. of 6 M 
H 2 SO 4 and HNO 3 , dil., and decolorize the Fe f 3 with a few drops of coned. H 3 PO 4 . 
Add several drops of 5 or 10% H 2 O 2 . A yellow to brown color indicates Ti IV 
or V v . Add a few mg of solid NaF to decolorize the titanium peroxide. The 
residual color is due to vanadium. The test will detect 500 ppm V. 

Test 20-9. Heteropoly Acid Color Test for VO$. Put a speck of 
NH 4 VO 3 on a spot plate, add 2 drops of H 2 O, 1 drop of 6 M H 3 PO 4 , and 1 drop 
of 10% Na 2 WO 4 soln. The formation of a yellow-orange color constitutes a 
test for V and is probably due to a series of vanadotungstic and vanadophos- 
phoric heteropoly acids such as H 2 WO 3 (VO 3 ), HPO 2 (VO 3 ) 2 , etc. Alternately 
this test may be used to detect WO^ 2 or PO^ 3 . 

Test 20-10. Vanadium and Its Minerals. Fuse one part of powd. 
mineral with three parts of KNO 3 for 5 min., then extract the potassium vanadate 
with hot H 2 O. Add BaCl 2 soln. to the clear extract as long as pptn. occurs. 
Collect the barium vanadate and decompose it with H 2 SO 4 . Centrifuge. 
Discard the residue of BaSO 4 and saturate the centrate with NH 4 C1. A ppt. 
of (NH 4 ) 3 VO, slowly forms. Dissolve a little in H 2 O, and try tests 20-8 and 
20-1 1 . (Sec reference 8, p. 356.) 

Test 20-11. Strychnine Test for VO^. A test which uses strychnine as 
a reagent for vanadate (or vice versa) is as follows: put a drop of ammonium 
vanadate on a spot plate and add a drop of coned. H 2 SO 4 and a crystal of a 
strychnine salt. A violet to rose color is a positive test. The cause of the color 
is not definitely known. Determine, roughly, the concn. limit. 

Caution! Strychnine is poisonous. 

Test 20-12. Precipitation of Be(OH) 2 . Caution! Beryllium metal, 
particularly in the finely divided state, and beryllium salts are poisonous. Dissolve 
a few mg of a beryllium salt, such as BeSO 4 -4H 2 O, in a ml of H 2 O and add a 
drop of 6 M NaOH. The white gelatinous ppt. is beryllium hydroxide. Add, 
with shaking, additional drops of NaOH to just dissolve the ppt., giving the 
beryllate ion, Be(OH) 4 2 , in soln. Save this soln. for test 20-13. (Alternately 
one could have shown that Be(OH) 2 is not sol. in excess NH 4 OH since no 
ammonia complex forms, but that in boiling 5% NaHCO 3 soln., Be(OH) 2 is 
sol., in contrast to A1(OH) 3 . This test effects a separation of the two metals 
in group 3 if Be 11 is part of a general unknown.) 

Test 20-13. Reprecipitation from Be(OH)^. Place the tube contg. 
the soln. from test 20-12 in a boiling water bath for 8-10 min. and note the 
gradual pptn. of /?-Be(OH) 2 . This is only about 1/25 as soluble as Be(OH) 2 
originally prepd. The test also distinguishes Be 11 from A1+ 3 , since the latter's 
hydroxide does not ppt. from AKOH)^ in the same test. 

Test 20-14. Quinalizarin Method for Be in Minerals. If a beryllium 
mineral is available, powder a little and fuse it (hood) with about four times its 
weight of potassium bifluoride, KHF 2 , using the technique described in test 20-5. 
Put 1 drop of clear aq extract on a spot plate, a drop of a known Be salt soln. in 


another depression on the plate, and a drop of water in a third depression. To 
each add a drop of quinalizarin (50 mg in 100 ml of 10% NH 4 OH) and note the 
color changes which indicate a positive reaction. To each add a drop of 
bromine water. Only the blank will be bleached. The concn. limit is 3 ppm. 
See reference 5, p. 356. 

Test 20-15. Identification of Ti Metal, (a) Examine a piece of Ti 
sheet. Note that its sharp edge can be used to mark glassware, since it is just 
hard enough to scratch and leave a metallic coating in the scratch. 

(/>) Dissolve a tiny piece of Ti in a ml of dil. H 2 SO 4 kept hot in a bath for 
15-30 min. Note the violet color of Ti m . Put a few drops in another tube, 
add a drop of coned. H 3 PO 4 and a drop or two of 3 % H 2 O 2 . Ti lv is the oxidation 
product and it forms the yellow to red pertitanic acid, H 2 TiO 4 (?). Add a few 
mg of NaF and note the effect as the stable complex TiF G " 2 forms. Cr, Mo, 
and V interfere by giving peroxide colors but do not give the fluoride reaction. 
The concn. limit is 30 ppm. 

(c) To another portion of soln. from (b), add a few drops HNO :l and heat to 
oxidize Ti lu . Now make the soln. basic with 6 A/NaOH and see if the hydroxide 
ppt. is amphoteric. 

Test 20-16. Uranyl Ferrocyanide. In the range 8-40 ppm. of U, 
ferrocyanide will give a red color with uranyl ion in mildly acidic soln., and the 
color is suitable to colorimetry. (In higher concns., a reddish ppt. of uranyl 
ferrocyanide forms.) 

Dissolve a few mg of a uranyl salt such as UO 2 Ac 2 -2H 2 O in 1 ml of H 2 O and 
divide the soln. between two tubes. To one add a few drops of K :j [Fe(CN) 6 ] 
and to the other a few drops of K 4 [Fe(CN) fi ]. The color in the latter reagent is 
due to (UO 2 ) 2 [Fe(CN) 6 ], or if excess reagent is present, the precipitate formed is 
(UO 2 )K 2 [Fe(CN) fi ]. (This reaction is sometimes used to indicate the end point 
in titrations with ferrocyanide.) Many metallic ions, particularity Fe +3 and 
Cu +2 interfere by giving dark colored ferrocyanides. 

Test 20 17. Peroxide Precipitation of 17 V1 . To 10 drops of 
soln. add 6 drops of 10% H 2 O 2 . A yellow ppt. of peruranic acid appears. The 
reaction probably is 

UO,/ 2 + H 2 O 2 + 2H 2 = 2H f + UO 2 (OOH)(OH) H 2 O 

Uranium may be quant, separated using this method, followed by freezing the 
mixture, then thawing and filtering off the peroxy acid hydrate at ice temp. The 
acid is sol. in alkalis from which one obtains sol., orange colored peroxide 
uranates, as Na 2 O 2 -UO 4 . 
For uranium in minerals, see special experiment 4, part 3(6). 

Test 20-18. Formation of ZrO(H 2 PO 4 ) 2 . Dissolve a few mg of 
ZrOCI 2 - 8H 2 O in 15 drops of H 2 O. Add 5 drops of dil. HNO 3 , heat to boiling, 
add 2 drops of coned. H 3 PO 4 , and observe the gelatinous ppt. of zirconyl dihydro- 
genphosphatc. This material is only slightly sol. in dil. nitric acid and the method 


is capable of detecting small amounts of Zr in complex mixtures. Apparently 
there are no common interferences to this test except Ti l v , which can be kept in 
soln. with 10% H 2 O 2 . 

Test 20 19. Colorimetric Test for Zirconium. Dissolve a few mg of 
a-nitroso-/y*naphthol in 1 nil of eth)l alcohol and to this add several drops of 
ZrO* 2 solution. A reddish color or ppt. similar to that obtained with Co+ 2 
and the same reagent results. Alternately the org. soln. may be spotted on 
filter paper or a spot plate and drops of zirconium soln. added to that. The 
concn. limit is about 10 ppm zirconium. 

Test 20-20. K.JZrF t J from Zircon, (a) Fuse one part of powd. zircon 
mineral with three parts of KHF 2 in a Pt or Ni dish. When cool, grind the 
mass and extract with hot H,O, contg. a few drops of HP. Decant into an 
inert container like a rubber beaker, or filter into it via a rubber funnel. Upon 
standing, potassium hexafluorozirconate crystallizes. 

Test 20-21. Precipitation of Lithium as the Phosphate. To 1 ml of 

Li+ soln., add 6 drops of (NH 4 ) 2 HPO t and heal the tube in a water bath. When 
hot, add 3 drops of dil. NH,OH and note the gradual accumulation of white, 
crystalline Li s PO 4 . In one quant, method for Li, this residue is filtered, dried, 
and weighed. Would any cations from group 5 (where Li ' is normally analyzed) 
interfere with the procedure? If so, how could interference be obviated? 

Test 20-22. Solubility of Lid. Put a few crystals of LiCI in each of 
several tubes and determine its solub.' in several org. solvents. Using the best 
solvent, try it on KC1 and NaCl, and suggest a means of separating LiCI from 
other chlorides in group 5. 

Test 20 23. The Potassium Ferric Periodate Method for Li ( . To 

a half ml of LiCI soln., add 1 ml of 1 M KOH and place the tube in a hot-water 
bath. In another tube put 2 ml of potassium ferric penodate reagent (see 
appendix A 13 for prcpn.) and heat this also in the bath. When hot, pour the 
latter into the lithium soln. and allow the mixt. to remain hot for 5 mm. The 
formation of light yellow lithium potassium ferric paraperiodate, LiKFe(IO r ,), 
is evidence that Li f was in the soln. (This method is adaptable to the quant, 
anal, of lithium in an indirect way. The ppt. is segregated, washed with 1 M 
KOH, dissolved in 1 M HC1, and to the acidic soln. is added an excess of 2 M 
KSCN. The concn. of red Fe(SCN) t2 is determined colorimctrically and since 
lithium is stoichiometrically related to ferric iron in the penodate ppt., one is 
able to relate color density of Fe(SCN) f a to [Li f ] in the orig. soln. See reference 
6, Chapter 12.) 

Test 20-24. Lithium by the Triple Acetate Method. Add a ml of 

zinc uranyl acetate (Na f reagent, test 19-6) to a crystal of a sol. Li 4 salt, shake 
the mixt., and note the pptn. of yellow LiZn(UO 2 );,Ac 9 -6H 2 O. This method 
is used in quant, gravimetry for lithium. 


Analysis of an Unknown 

An instructive exercise at this point is to give the student a sample of 
salt, oxide, pure metal, alloy steel, or other mixture of the less familiar 
elements for analysis. There is enough experimental method given for 
simple samples. The student should first do problem 2 below. 


1. The diagrams* p. 357 can be largely constructed from the discussion on 
molybdenum. Prepare similar diagrams for the other seven elements described 
in the chapter. 

2. Consider having a solution containing equal quantities of WO^ 2 , MoO^f 2 , 
VO.f , TiO+ 2 , ZrO+ 2 , BeOH+, UOf 2 , and Li + . Prepare a flow sheet from infor- 
mation in the chapter and supplementary reading, as necessary, for the partial 
separation and identification of the ions. 

3. Consider having a solution containing the ions of problem 2, plus all the 
other cations described for analysis in this text. Prepare a flow sheet showing 
separation of every ion, and confirmatory tests where desirable. 


1. T. Chao and J. Yang, /. Chem. Educ., 25, 388 (1948). (Tungsten) 

2. W. B. Blumenthal, /. Chem. Educ., 26, 472 (1949). (Zirconium) 

3. S. G. Sjoberg, J. Chem. Educ., 28, 294 (1951). (Vanadium) 

4. J. H. Yoe and A. R. Armstrong, Anal. Chem., 19, 100 (1947). (Titanium) 

5. F. Feigl and A. Schaeffer, Anal. Chem., 23, 351 (1951). (Beryllium) 

6. F. N. Ward, Anal. Chem., 23, 778 (1951). (Molybdenum) 

7. E. R. Caley and G. A. Simmons, Anal. Chem., 25, 1386 (1953). (Lithium) 

8. A. D. Weeks and M. E. Thompson, Geol. Survey Bull., 1009-B (1954). (Uranium, 

9. C. J. Rodden, Analytical Chemistry of the Manhatten Project, McGraw-Hill, 
New York (1950). 

10. B. S. Hopkins, Chapters in the Chemistry of the Less Familiar Elements, Stipes, 
Champaign, (1939). 

11. J. J. Katz and E. Rabinowitch, The Chemistry of Uranium, McGraw-Hill, New 
York (1951). 

12. A. A. Noyes and W. C. Bray, A System of Qualitative Analysis for the Rare 
Elements, Macmillan, New York (1927). 

Numerous publications are available from the U. S. Bureau of Mines, Atomic 
Energy Commission, Foote Mineral Co., Zirconium Corp. of Amer., Metals Disintegrat- 
ing Co., Titanium Metals Corp. of Amer., Mai lory-Sharon Titanium Corp., Metal 
Hydrides Inc., Fansteel Metallurgical Corp., Molybdenum Corp. of Amer., Rem-Cru 
Titanium Vacuum Metals Corp., Republic Steel Corp., Climax Molybdenum Corp., 
and others. 

* Adapted by permission from data furnished by Climax Molybdenum Co., 500 
Fifth Ave., New York 36, N.Y. 



Molybdenite Ore 
0.3% Mo) 



MoS 2 


MoS 2 


Na 2 MoO 4 


MoO 3 

NH 4 OH 

(NH 4 ) 6 Mo 7 O 24 -4H 2 O 




FIG. 20-2. Molybdenum process metallurgy. 

NH 4 OH Reduce 


Mo0 3 



A, C 

FIG. 20-3. Reactions of metallic molybdenum and some of its compounds. 






The student who has studied Chapters 14-20* should by this time have 
a good idea concerning the direction that analysis of a general cation 
unknown will take. The flow sheet in Chapter 14 is a summary of this. 
Generally speaking, the sample is separated into groups in the order given 
and the individual groups analyzed according to their respective flow sheets. 
Student samples are assumed to be free of serious interferences such as 
unknown complexing agents and/or elements nbt described for separation, 
though such complications arc the rule, in samples the industrial laboratory 
gets. Three common interferences are organic materials, borates, and 
phosphates, and the paragraphs following illustrate typical methods used 
in their presence. 

The instructor Mi/I inform the class ij the following precautions need be 

If the sample is already in solution, start with 1-3 ml, as directed by the 
instructor, and begin analysis with group I, p. 243. 

If the sample contains some solid material, the solid is probably a com- 
bination of group 1 chlorides and/or group 2 hydrolysis products of Bi, 
Sb, and Sn. The solid may be separated by centrifuging and treated as 
an individual sample. Directions for solid mixtures are given below. 

Detection and Removal of Organic Compounds 

Evaporate a small portion of unknown slowly in a crucible and bake 

* The "Jess familiar" elements of Chapter 20 will not be included in this discussion. 



the residue. A charring may indicate the presence of carbon-containing 
compounds, though some inorganic compounds may become dark also, as 
for example cupric nitrate which is converted to black cupric oxide by 
heat. It is assumed that explosive mixtures, which might include chlorates 
and perchlorates, are not present. 

Organic matter is removed from samples prior to inorganic analysis 
because it may color the solution, complex or precipitate ions, or use up 
carefully measured reagents that otherwise would be reacting with specific 
ions. Hot acid digestion is the usual method of decomposition. In the 
hood using a 20-30-ml crucible, heat a mixture of 2 ml ot unknown with 
5 drops of concentrated H 2 SO 4 and 10 drops of concentrated HNO 3 . 
Evaporate to a volume of about 5 drops, whereupon white fumes of SO 3 
will be visible. Further HNO 3 addition and heating may be necessary if 
decomposition appears incomplete. This treatment not only rids the 
solution of organic compounds in general but specifically oxidizes and 
destroys the common tartrate and oxalate ions and the NHJ ion, and will 
distill out all low-boiling acids such as HAc and HF which otherwise 
would form complexes with Fe +3 , etc. 

Dilute the cooled mixture dropwise with 2 ml of H 2 O, transfer it to a 
tube, and centrifuge. Any residue that was not present before will be 
the sulfates of Pb, Ba, and possibly Ca and Sr. After washing, this solid 
may be put into solution by boiling with 2ml of 1.5 M Na 2 CO 3 ; the 
carbonates are then solubilized ' in HAc and tested separately. The 
centrate from above is analyzed as a general unknown. 

Detection and Removal of Borates 

Borates precipitate certain group 3 and 4 ions and can confuse analysis 
because of that action. Borates are detected by tests 22-23, 22-25, 
and 22-26. 

To the centrate ajter group 2 removal, add 10 drops of concentrated HC1 
and 10 drops of methyl alcohol. Boil the solution under the hood and 
resupply HC1 and alcohol several times until the alcohol vapors burn 
light blue, as contrasted to the previous green due to methyl borate, 
B(OCH 3 ) 3 . Let the excess alcohol burn off. The solution is then carried 
to the group 3 procedure, and since all the CH 3 CSNH 2 has now hydrolyzed 
it will be necessary to add more (NH 4 ) 2 S than provided for in the group 3 
description which assumes the solution for testing is directly from group 2 

If borate is removed prior to group 2 analysis in the manner described 
above, Sb, As, and Sn are largely lost as their volatile chlorides, so the 
group 2 procedure is run first. 


Detection and Removal of Phosphates 

Orthophosphate precipitates tripositive ions in group 3 at high /?H or 
complexes them as soluble, stable acid phosphates at low pH. The 
procedure is to test the group 3 solution for Fe, then to precipitate PO^ 3 
as FePO 4 by adding Fe+ 3 . Other phosphates can also cause trouble* 
but are converted with hot HNO 3 to the orthophosphate. See test 22-16 
for POj 3 detection. 

Boil the group 2 centrate to drive out all the H 2 S. Add 3 drops of 
concentrated HNO 3 and reboil. Test 1 drop of this solution for Fe^ 3 
with SCN~. Report Fe present in the unknown if the test is positive. 
To the main portion of solution add 6 drops of NH 4 Ac, then 15 M 
NH 4 OH until basic, 6 M HAc until just acidic, and then 2 drops more. 
Jf Cr+ 3 , Fe+ 3 , and/or A1+ 3 are present they will precipitate as phosphates; 
if not present no precipitate will form. Add 0.1 M Fe+ 3 dropwise until 
no further FePO 4 precipitates, centrifuge, and note that the supernatant 
liquid is the Fe 4 * 3 color, indicating iron in excess of that needed for complete 
PO 4 3 precipitation. Adjust the volume to about 4 ml and boil the solution. 
After several minutes centrifuge. The residue may contain hydroxides 
(due to hydrolysis) and phosphates of iron, chromium, and aluminum, 
while the centrate contains the other group 3 ions as well as those of groups 
4 and S.f Fe(OH) 3 is separated from the residue by solubilizing the 
latter with excess 6 M NaOH plus heating with a few drops of 3 % H 2 O 2 
and centrifuging. The centrate from the Fe(OH) 3 removal may contain 
CrO 4 2 (yellow) and A1(OH) 4 (colorless). This mixture is readily handled 
by methods given in Chapter 17. The centrate from the original basic 
acetate separation is analyzed for other group 3 ions by starting at the 
beginning of the group 3 procedure. 

Solid Samples 

Several cases may arise if the sample "as received" is not already in 

1. The Solid is Water Soluble. If this is observed, no preparation 
for general cation analysis is necessary. 

Dissolve about 75-100 mg of solid in 5 ml of water and using a 1-2 ml 
sample, go through the five cation group procedures. Other portions 
of solid and liquid sample are then used for special tests as seem advisable. 

* Metaphosphate, PO^~, precipitates Ba +2 and pyrophosphate, P 2 O^~ 4 , precipitates 
Mn+ 2 . 

t This method is called the basic acetate separation. The residue may also contain 
V, W, Zr, and Ti as hydroxides, phosphates, or basic acetates, and will contain the rare 
earths as phosphates if those ions are present. If PO^" 3 is not present, the rare earths 
(including uranium) are retained in the centrate. 



More concentrated sample solutions are not needed because the text 
directions and equipment provide for handling small amounts. 

2. The Solid Appears Nonmetallic and is Not Water Soluble. 
Minerals, ores, and nonsoluble salts fall within this classification. A 
first observation to make is the sample's color for a hint to its identity. 
The accompanying table lists a few typical cases. 



Compound Type 

Metals Possibly Present 




As, Sb, Sn, Al, Zn, Mg, Si, Ti, Zr, Be 



Pb, Ba, Sr, Ca, Mg 
Pb, Ba, Ca 



Ca, Mg 

Bi, Cu, Fe, Ni, Co, Mn, V, U 



Ag, Pb, Hg, Cu, Sb, Fe, Ni, Co, Mo, V 
Pb, Bi, Cd, Fe, Mn 

V, Pt 





Pb, Hg, Cu, Fe 

Cd, As, Sn 





C, Ni 


Other salts 

Many insoluble salts. See solubility 
in handbooks and this text. 


* Colors of natural minerals vary widely and are sometimes different from the 
compounds prepared in the lab; solubility of the former is apt to be lower as well. 
See mineralogy references following special experiment 4. 

Use a few milligrams of sample to determine its solubility in the following 
solvents used both dilute and concentrated, hot and cold: HNO 3 , HC1, 
and aqua regia (one part of HNO 3 to three parts of HC1). If HC1 or 
aqua regia is used, a white precipitate may indicate the presence of group 1 
ions. Some common substances which aqua regia will not dissolve are 
the sulfates of Pb and of the group 4 metals, CaF 2 , the Ag halides, C, S, 
A1 2 O 3 , SiO 2 , and most silicates. The acid solution will probably contain 
some ions from the sample that may be partly soluble and is analyzed as a 
general unknown after boiling down to decompose excess aqua regia. 

The acid-insoluble residue is centrifuged, washed with water several 
times, and boiled with 2 ml of 1.5 M Na 2 CO 3 to transpose it to the carbon- 
ates of metals that are then soluble in dilute HNO 3 . This treatment will 


not solubilize the last five mentioned substances, however, and further, 
more drastic methods are needed. The silver halides are solubilized by 
heating with a mixture of 3 M H 2 SO 4 and granular zinc, the metal reducing 
Ag* to Ag, which then can be dissolved in HNO :t and tested as in group 1. 
The remaining insoluble materials can be fused in a nickel crucible with a 
Na 2 CO 3 -K 2 CO3 mixture or with any of these: Na 2 O 2 , K 2 S 2 O 7 , or KHSO 4 . 
The cooled melts are taken up in dilute HNO 3 and the solution treated as a 
general unknown. (Directions for fusions arc given in books dealing 
with ore and mineral analyses.) Fusions can give high oxidation states 
like MnO 4 and CrO^ 2 . MnO 4 will oxidize HC1 in group 1 precipitations 
to H 2 O + C1 2 , and MnO 4 or Cr 2 O 7 2 will oxidize H 2 S to H 2 O + S under 
group 2 precipitation conditions. The color changes (review Chapter 17) 
will serve to indicate what is taking place and more of the precipitation 
reagents must be added to compensate for loss. 

3. The Solid is Metallic. HNO 3 is the best solvent for most metallic 
samples, though Cr and Al are only partially soluble due to development 
of impervious oxide coatings. In addition. Sb and Sn form insoluble 
hydrated oxides. HSb(OH) fi or Sb 2 O s H 2 O and H 2 SnO.< or SnO 2 -H 2 O. 
Other than these cases, HNO 3 has the advantages of (a} yielding soluble 
nitrates with most metals and (/>) oxidizing elements like P. As. S. and 
Hg (also Sb and Sn as above) to PO 4 3 , AsOf 3 . SO, 2 , and Hg i2 . which in 
the presence of hot HC1 would be lost bv the volatility of PHo, AsH., 
H 2 S, SbH 3 , Hg, and SnCl 4 . 

A metallic sample not dissolving completely in HNO^ after prolonged 
heating and further additions of acid, probably contains Sn and /or Sb 
among the common elements. This residue of hydrous oxide(s) can be 
solubilized by fusing in an iron dish with a mixture of Na 2 CO ;l -f S, 
giving SnSa 2 and SbS 4 3 . which are extracted with hot H 2 O. That solution 
is treated as a group 2B unknown while the centrate from the Sn-Sb 
separation is boiled down to expel most of the HNO ? . then diluted and 
handled as a general unknown. Group 2 will need extra time to precipi- 
tate because As^ present oxidizes HS until reduced to As m (or NH,I 
can be employed to effect the reduction). When the FCOH separation 
is then made the centrate may be combined with the extract from the 
carbonate-sulfur fusion and analyzed as a 2B sample.* The analysis is 
continued through group 3 and the centrate. from its separation, should 
be colorless and will generally include only Mg 12 as a possibility. 

If the alloy is not appreciably soluble in dilute HNO 3 , aqua regia 

* The centrate for groups 3 through 5 should be tested for POp 3 , which may come 
from the oxidation of phosphides, and, if found, the group 3 procedure is altered to 
include the basic acetate separation already described. Student samples will rarely 
need this, though P is a common element found in small amounts in steels and bronzes. 


should be tried and if that fails, one may try a mixture of concentrated 
HC1 plus a few drops of bromine or saturated Br 2 water. (Caution! 
Only the laboratory instructor or stockroom man should handle highly 
corrosive liquid Br 2 . Any accidental contact with the skin must be counter- 
acted bv immediate application of sodium thiosulfate solution.) In any 
case, excess solvent is evaporated under the hood, the residue of salts 
redissolved in water, and the analysis begun with group 1. A white- 
yellow residue may be AgBr, which is dissolved by the method described 
for AgCl on p. 362. 

Samples for beginners are usually soluble in the common solvents and 
relatively free of complications such as the presence of unfamiliar elements. 
If the samples are commercial alloys, the student should make use of 
library references in the metallurgical field to get clues on the composition, 
physical appearance, and other characteristics of metals whose identity is 
sought. Such samples present the added problem of containing a few 
metals in major proportions and others in small amounts. Only careful 
work will discover the latter. 

Dry Ignitions 

Heating a small portion of solid sample on a wire gauze or in a test 
tube made from small-diameter glass tubing, sometimes gives the analyst 
valuable information on the composition of the material with a minimum 
amount of effort. Any clues he gets here will suggest spot tests for 
specific ions or elements. 

1. The sample is metallic and burns brightly: Mg and its alloys. 

2. The sample is metallic and melts at a relatively low temperature 
perhaps even in hot H 2 O: Bi, Pb, Sn, Sb, Cd, and their alloys. 

3. The sample is nonmetallic. A gas is released and color may change. 

(a) H 2 O: hydrates, NH+ salts, bicarbonates, hydroxides. If the 
steam is basic to litmus, NHJ salts are indicated; if the reaction is 
acidic, strong acid radicals are indicated. Odor may help identifica- 
tion, as in the case of NH 3 . 

(b) O 2 , as tested by glowing splint: nitrates. 

(c) Nitrogen oxides : nitrites and possibly nitrates. 

(d) CO 2 , as tested by lime water: some carbonates, bicarbonates, 
organic matter. 

(e) CO (POISON!), light blue flame when burned: oxalates. 
(/) ^2' purple gas: some iodides. 

(g) C1 2 , yellow gas: some chlorides. 

(h) Br 2 . brown gas : some bromides. 

(/) H 2 S, odor: some sulfides. 

(/) SO 2 , odor: sulfites, thiosulfates, some sulfates. 

7-7,-g & 







O a 






$ e St 




I s 



4. The sample is nonmetallic and changes color when heated and may 
change again when cooled. 

(a) Chars or burns : organic matter. 

(b) Simply turns black : some Cu, Mn, Ni, Co salts. 

(c) Yellowish when hot, fades when cool: ZnO, SnO 2 , some Zn 

(d) Yellow hot and cold, tends to fuse into the glass: some Pb 

5. Sample may or may not appear metallic but gives a sublimate when 

(d) Black to grey sublimate: some iodides, and mercury and arsenic 
compounds. The mercury coalesces in droplets, the arsenic has a 
garlic odor (POISON!). 

(b) White sublimate: some Hg and NHJ halides and arsenic and 
antimony oxides (POISON!). 

Industrial Analysis of Metallic Samples 

Quantitative analysis of metal samples in industry is of prime importance 
since physical properties of alloys are altered by composition changes 
even of minor sorts; hence rapid, accurate, and sensitive methods are 
needed as quality controls. The spectrograph, which is most often used 
here, meets this need admirably. It is an instrument in which the sample 
is burned in a high temperature arc, the light from the combustion con- 
sisting of spectra of the metals in the sample. (See special experiment 5.) 
The spectra are broken up into their component lines by a diffraction 
grating, a section of metal surface on which is ruled some 10-35,000 
lines to the inch, and the spectral lines photographed. Since each 
metallic element has its own characteristic lines, the photo is analyzed 
first qualitatively for the sample composition by the number and position 
of the lines. The density of important reference lines for each element is 
then determined with an instrument called a demitometer, which relates 
density of the photographed lines to the quantity of each element present. 
The automatic recording spectrograph (~ $40,000) when set up, for 
instance, to routinely analyze brass samples, is capable of analyzing a 
10-15 element sample every 5-10 minutes and rendering a printed report. 
A recently developed X-ray fluorescence spectrometer does the same job 
without destroying the sample. Using it one can even monitor a flowing 
stream of substances in solution. The stream is irradiated by high intensity 
X-rays, causing a fluorescence from excitation of atoms. The wave length 
of the radiation is a qualitative indication of an element's presence, the 
radiation intensity at a particular wave length is the quantitative measure 



of the element. Potassium and all heavier atoms can be measured by this 

Qualitative analysis of metal samples in industry is often still done by 
simple laboratory methods although test instrumentation increases every 
year. Inorganic alloying elements or impurities like S, P, C, N, and B 
are largely determined by old, familiar wet chemical methods. Most 
testing is done for the purposes of meeting specifications, identifying 
mixed up stock, and sorting scrap for salvage. 

Methods for metal identification in industry are noted for their simplicity 
and speed, because only a limited number of possibilities are encountered 
in a given plant, and a single test may be sufficient in some cases. For 
example, the theoretician if asked to differentiate between type 310 
stainless (20% Ni, 25% Cr, a little Mn and C, balance Fe), and type 446 
stainless (28 % Cr, a little C, balance Fe) might spend several hours cutting 
off samples, dissolving them in acid, neutralizing, eliminating iron and 
chromium, and testing for nickel and manganese. The practical man 
might have a magnet in his pocket and know that 446 is magnetic and 
310 is not. In another instance the machinist might in a few seconds 
identify a bar of iron by the type of sparks it gives on grinding with an 
emery wheel (the spark test), while the chemist fresh from the classroom 
is still trying to remember where he stored his qualitative analysis book so 
he can look up a few clues on how to begin. Partial identification of 
metals by nondestructive means are common. A recent advance in 
qualitative testing for nonferrous metals is the Sigma test (Magnatest) in 
which an electronic conductivity meter is used to induce a high frequency 
eddy current in the sample for pickup on a small probe in contact with the 



Specific Gravity 

Metals and Their Alloys 


Very heavy 



Be, Al, Mg. (Dowmetal, Alcoa, etc.) 
Co, Cu, Fe, Ni, Sn, Zn, Sb, Cd, Nb, Mn, Ti, V, 
Zr. (steels, brasses, bronzes, hastelloys, 
Pb, Ag, Bi, Mo. (solder, type metal, etc.) 
Au, Ir, Hg, Os, Pd, Pt, Re, Rh, Ru, Ta, W, U. 

(jewelry, catalysts, etc.) 

* Partially adapted from "Rapid Identification (Spot Testing) of Some Metals and 
Alloys," published by The International Nickel Co., Inc., New York 5, N.Y., and 
printed here by their permission. See also Appendix A23. 



metal. Conductivity measuied in this way is a function primarily of 
alloy composition. Industrial qualitative testing ol metals is also done 
by the elecrrographie method (special experiment 13) and by some spot 
testing chemical procedures following observations on physical features 
of the sample which suggest or rule out various methods. The latter are 
perhaps of most interest to chemists. 

Specific gravity is an obvious simple screening test for metals. The 
following table is self-explanatory in this regard. 

Having classified a sample roughly by density and knowing further that 
cnly a limited number of commercial alloys would be used, one can devise 
quick tests for his own problems. 

A Light Alloy Problem 

Suppose one has a light metal sample and knows that only aluminums 
2S, 3S, and 75S, and Dowmetals M and C are ordered in the factory. 
From a study of the compositions guaranteed by the manufacturers, one 
is able with a little ingenuity to arrive at a system lor handling the matter, 



Per Cent Composition 










99. 5 -f- 














Dowmetal M 



Dowmetal C 




* See note to 'fable 21-2. 

as shown in Table 21-3. One may draw upon facts like the following for 
differentiation of alloys in the table: 

(a) The Al-based alloys are known to have an oxide coating which the 
Mg alloys do not have, hence a dilute Ag f solution is rapidly reduced to 
black metallic Ag only on the Mg surfaces. 


































































































































































































































































































































































































^ - 






































































































x 3 










































































(b) Mn may be found by putting a drop of NaOH on the specimen to 
dissolve the oxide coating, followed by excess concentrated HNO :i , 
H 3 PO 4 , and NaBiO 3 to produce the purple MnOj color. 

(c) A dilute CdSO 4 -NaCl-HCl solution (5 g CdSO 4 - 3/8H 2 O, 3 g NaCl, 
5 ml 14 M HC1, dilute to 100 ml) gives a dark spot of cadmium metal if 
zinc is present. 

(d) Properly prepared, the aluminon test 17-6 for Al will differentiate 
the Dowmetals if (c) fails. 

A Stainless Steel Problem 

There are many stainless steels, all containing Fe, Cr, Ni, C, and minor 
elements. To distinguish among them might appear unfeasible without 
spectrographic analysis, but the flow sheet given in Fig. 21-1 and identifica- 
tion tests* show how the job is accomplished with a minimum of effort 
using ordinary laboratory tests. 


Preparation of Samples. All specimens must be free from scale . . . 
which may be removed by pickling or grinding. In making a Spot Test, always 
grind a fresh area before applying the solution. 

Nitric Acid Test. This testing solution is made by mixing one part of 

concentrated nitric acid and one part of water. A drop of the solution is placed 

on a freshly ground snot on the snecimen. Attack is denoted by a boiling 
action or gas evolution from the drop. 

Magnet Test. Austenitic stainless steels fthe 18-8 types) are nonmagnetic 
in the annealed condition, but become slightly magnetic when cold-worked. 
This slieht magnetism can easily be distinguished from the stronger magnetism 
of the straight chrome types. 

Spark Test. A spark test is made to separate stainless types 420 and 440 
from nonferrous alloys and ordinary iron and steel as nitric acid, under certain 
conditions, may attack annealed specimens of some stainless types. These 
stainless steels are identified by the characteristic short "chrome" spark. 

Hardness Test. Specimens are heated to 1850 F. and oil quenched. 
Various carbon percentage ranges will be noted by different hardnesses as 
quenched. The non-hardening types will show low hardness as oil quenched 
from 1 850 F. 

Muriatic Acid Tests. Specimens are placed in a solution of Muriatic 

* Both reprinted by permission of The Carpenter Steel Co., Reading, Pa. Re 
refers to the Rockwell C scale of hardness. If these values are furnished, stainless 
samples make interesting unknowns. Standard samples should be available for student 


Acid* (one part water, one part Muriatic Acid) at a temperature of 180-190 F. 
A fresh solution should be used for each of the following tests : 

(a) After a five minute immersion, type 303 will be coated with a heavy black 

(k) Within two minutes after immersion, Type 316 can be identified by the 
very slight acid attack as compared with the active attack and evolution of gas 
on types 302, 321, and 347. 

(c ) Immerse specimens for at least fifteen minutes. Type 443 can be identified 
by a brownish -colored smudge. 

(d) Immerse specimens in a separate container for five minutes, then remove 
them from the solution. A pungent garlic-like odor (Caution) will be 
detected on types 416 (Se) and 440F. This characteristic odor 6f hydrogen 
selenide gas will not be present in pickling types 410 and 440. 

Sulphur Spot Test. Three drops of sulphuric acid solution (one part 
sulphuric acid, three parts water) are placed on a newly ground spot on the 
specimen and allowed to react for one minute. One drop of a 5% solution of 
lead acetate in water is then added to the acid drop and allowed to react for 
15 seconds. The spot is then washed with water and examined. A positive 
test for sulphur (stainless types 416 (S), 420F, and 430F) is the presence of a 
black sulphide deposit. 

Nickel Spot Test. Three solutions are required for this test: 

Solution no. 1 : Mix 125 cc of water, 100 cc concentrated nitric acid, and 25 cc 
of 85 % phosphoric acid. 

Solution no. 2: One part water and one part muriatic acid. 

Solution no. 3: Dissolve one g. of dimethylglyoxime in 60 cc of glacial acetic 
acid. To this solution add another solution, prepared by dissolving 10 g of 
ammonium acetate in 30 cc of ammonium hydroxide. 

Procedure: Place one drop of solution no. 1 on a newly ground spot on the 
specimen. Allow it to react for 30 seconds and then add one drop of solution 
no. 2. This mixture is allowed to react for an additional 30 seconds and is then 
absorbed with whke blotting or filter paper. One drop of solution no. 3 is 
then placed on the moist spot of the paper. The formation of a red or pink 
coloration in the spot denotes the presence of nickel. 

Stabilization Test. Test specimens are heated to 1250 F for two hours 
and then cooled to room temperature in air. The specimens are then placed 
in a cold solution of three parts nitric acid, one part hydrofluoric acid, and six 
parts water and left for one hour. Remove specimens from solution and wash 
with water. Stainless types 321 and 347 will show a very slight attack from the 
acid. Stainless types 302 and 304 will have rough, granular surfaces. 

The Practicing Quality Analyst 

Many procedures of the types shown above have and are being routinely 
composed by analysts working in industrial laboratories. The information 
* Technical grade coned. HO. 


they want is, as a rule, not found in texts and they are continually 
forced to plumb their training in fundamentals, their imagination, experi- 
ence, and inventiveness to produce answers to problems where no set 
approaches are known. It is evident that the more one knows of fields 
allied to qualitative testing, such as metallurgy, physics, and mathematics, 
the more valuable he is as an analyst. Almost anybody can follow cook- 
book directions or substitute numbers into formulas. But when no 
directions or formulas are given, only those people capable of doing 
original thinking and who are willing to put forth some study and effort 
go on from there. There is great satisfaction in being creative and 
chemistry offers opportunities for creativity in every direction. 


1. Draw a flow sheet for the separation of the light metal alloys in Table 
21-3 using the chemical methods described there. 

2. How could one distinguish between the members of the following pairs of 
materials by simple chemical and/or physical tests? Give two methods where 

(a) Plain C steel and stainless steel. 

(b) Brass and Cu. 

(c) Sea water and Nad solution of the same density. 

(d) U and Fe. 

(e) Al and Be. 

(f) A 10-10-10 fertilizer and a 10-0-10 fertilizer. 
(#) Tungsten carbide and iron carbide. 

(h) Hadfield steel and high C steel. 

(/) WandTi. 

(j) Hastelloy C (Ni, Fe, Mo) and Chromel C (Ni, Fe, Cr). 

3. An alloy has a specific gravity of 9.5 and melts at 200 C. What elements 
might be present ? 

4. You are asked to determine whether a piece of wire is pure tantalum or 
pure rhodium. From the literature, find one physical and one chemical test 
for doing the job in a hurry. 

5. One has three samples of pure metals: Pb, Mo, Bi. Of these, only one is 
attacked by HAc and of the two remaining, HNO 3 attacks only one. Explain. 

6. What methods could one use to distinguish between a brass (Cu-Sn-Zn) 
and a bronze (Cu-Sn) which have about the same color and density? 

7. (Library) Look up the spark method for testing steels and give a brief 
explanation and a few examples of observations on specific steels. 

8. Hot H 2 SO 4 is used to destroy organic matter in a sample and results in 
precipitating a white solid that may be SrSO 4 , PbSO 4 , and/or BaSO 4 . This is 


segregated. How does one proceed to put it back in solution and test for the 
metals ? Prepare a flow sheet for this. 

9. Why are Ba, Sr, Ca, Na, K, NH 4 , Hg, and As seldom (and some of these, 
never) found in engineering alloys ? 

10. (Library) Look up a few of the main facts concerning Ta and compare it 
to Ti. Repeat for Nb and compare to V. 

11. Refer to Appendix table A23. Suppose one had pieces of Hastelloy B, 
Hastelloy C, Hastelloy D, and Invar which needed sorting. Draw a flow sheet 
and implement it with directions for doing the job with minimum equipment. 

12. (a) One has an irregular piece of alloy to identify. Explain at least two 
ways in which its density can be determined. 

(b) Are there any metallic elements other than Cu and Au that are not silvery 

13. Explain the chemistry behind the steps in the Carpenter methods for 
stainless steel analysis. 

14. Jackson Slipshod says that hot acid digestion takes too long to decompose 
organic interferences and recommends heating the sample to 800 C and holding it 
there until the organic matter is gone. Is there anything wrong with this? 

15. A stainless steel is not attacked by HNO 3 , is magnetic, has a hardness, 
Re, of 8/24, and gives a black spot with dilute H 2 SO 4 followed by PbAc 2 . Which 
alloy is it ? Which elements has one tested ? How could the others be identified ? 
Prepare a complete flow sheet for testing all elements in the sample. 





The qualitative testing of negative ions is on the whole a less systematic 
job than the analysis of positive ions. The reason is that most anions 
are sufficiently different from each other so spot tests may be made 
directly upon complex mixtures without the necessity of much preliminary 
group or individual separations. Three reagents are used, however, to 
divide anions into four classes, and are employable in a first gross examina- 
tion. A negative test with any of these is of utility in that further testing 
for the anion group(s) shown thereby to be absent is not necessary. 
Following this, small portions of sample are tested for individual ions in 
groups which are present. If cation analysis has preceded anion analysis 
on a sample, solubility data are also of assistance in ruling out some 
anions. Specific directions for handling the sample for anion analysis 
is given in the last section of this chapter. 

Of the 30-40 common anions, 18 have been selected as being repre- 
sentative. They are divided into four groups: 

Anion Group 1: Decomposed by Dilute 

A. NO^ ^ 

v ^ 


H 2 O + HNO 3 + NO 

SO 2 + S + H 2 O 
C0 2 + H 2 

o 3 

Tests are made for gaseous products by odor or by characteristic reaction 



when bubbled into reagent solutions. It is evident that these ions cannot 
exist in strong acid media. (Other anions also decomposed under these 
conditions include hypochlorite, CIO", and cyanate, CNO~. These will 
not be considered.) See also paragraphs W, X, and Z concerning inter- 
ferences among ions of the various groups. 

Anion Group 2: Insoluble Ba+ 2 and for C0+ 2 Salts 
F. P07 3 \ ^ - Ba 3 (P0 4 ) 2)i 


I. BOr 




Ba(B0 2 ) 







When this combined residue is treated with hot, dilute HC1, only BaSO 4 
remains undissolved. The solution and residue are tested further. 
Adding Ca+ 2 and Ba+ 2 solutions separately gives a rough indication of 
ions present, as the composition of the precipitate above shows. 

Anion Group 3: Insoluble Ag+ Salts 

L. J Cl- 

VI. / Br 

N. 1- 

O. SCN- 

P. Fe(CN) 6 - 4 x 

The silver compounds are insoluble in dilute HNO 3 . The residue or 
original solution is tested further. 

Anion Group 4: Not Indicated by Other Group Reagents 

Q. N03j" 

R. ACT/I- 

One always must test for these, since no group reagent is available to rule 
them out by negative test. 

A. Nitrite Ion, NO 2 

Neither nitrous acid solutions nor most nitrites are stable. From 
the oxidation potentials below, one can deduce, for example, that nitrite 
is capable of reducing bromine and of oxidizing iodide: 

HN0 2 + H 2 O = NOg + 3H+ + 2*- = - 0.94 

NO + H 2 = HN0 2 + H+ + e~ E = - 1.00 


Because of the many oxidation states of nitrogen, reactions of nitrites 
can be complex, and other products than the above often appear. 
Nitrous acid may exist as either H O N=O or O-N=O, and two 


groups of isomeric compounds are related to these structures, as discussed 
in Chapter 4. Resonance in the ion may lead to attachment through an 
oxygen (unstable nitrites) or through the nitrogen (more stable nitro 

Nitrogen has oxidation states ( III) as ammonia NH 3 , ( II) as 
hydrazine H 2 N NH 2 , ( I) as hydroxylamine H 2 NOH, (I) as nitrous 
oxide N 2 O, (II) as nitric oxide NO, (III) as nitrogen trioxide N 2 O 3 (the 
anhydride of nitrous acid HNO 2 ), (IV) as nitrogen dioxide NO 2 , and (V) as 
nitrogen pentoxide N 2 O 5 (the anhydride of nitric acid). 

Except for some complex compounds, all nitrites are water soluble. 
(AgNO 2 solubility = 3.4 g/liter; see also test 22-4.) 

Nitrous acid may be prepared from NO 2 or N 2 O 3 and H 2 O or from a 
nitrite and strong acid. Many tests for nitrite are known. The most 
sensitive is the diazo reaction, given as example 11, Chapter 12, p. 195. 
Three others are (a) oxidation to NOg" with strong oxidants : 

5HNO 2 + 2MnC>4 + H+ = 2Mn+ 2 + 3H 2 O + 5NO 3 

(b) with Fe f2 in dilute acid to give a brown color due to nitrosyl iron(II) ion 

3HN0 2 + Fe+ 2 = H 2 O + H+ + NO 3 + Fe(NO)+ 2 

and (c) reduction to nitrogen by heating with an ammonium salt, or with 
acidified solutions of urea, thiourea, or sulfamic acid : 

N0 2 + NH+ = (NH 4 N0 2 ) = N 2 + H 2 O 
2HNO 2 + (NH 2 ) 2 CO = 2N 2 + CO 2 + 3H 2 O 

HN0 2 + (NH 2 ) 2 CS = JV 2 + HSCN + 2H 2 O 
HNO 2 + NH 2 SO 2 OH = N 2 + 2H+ + SOj 2 + H 2 O 

The last reaction is best for quantitative destruction of nitrite. 

One quantitative analytical method is reduction with base and active 
metal to NH 3 and titration of the ammonia. See test 22-47. Others are 
illustrated in tests 22-5 and 22-6. 

B. Sulfide Ion, S~ 2 

Sulfide sulfur has the lowest valence that element exhibits, therefore it 
can act only as a weak reducing agent : 

S~ 2 = S + 2e- = 0.51 



Many oxidizing agents are capable of causing sulfide decomposition in 
this way. H 2 S is somewhat soluble in water (~ 0.1 M at 20 C and 1 atm); 
its aqueous solution is called hydrosulfuric or hydrosulfic acid. By odor 
alone one can detect this poisonous gas at a concentration of about 0.8 
ppm in the air. Being a very weak diprotic acid, soluble sulfides are 
about as basic as comparable concentrations of strong bases, since 
hydrolysis to bisulfide is virtually complete. Little odor of H 2 S is detected 
over Na 2 S. See Chapters 7 and 10 for mathematical treatment of H 2 S 


Reactions (in order of decreasing [S~ 2 ]) 


MnS = Mn+ 2 + S~ 2 


x JO" 14 

FeS (a , = Fe+ 2 + S~ 2 


x 10~ 17 

ZnS = Zn+ 2 + S- 2 


x 10- 20 

NiS (a) = Ni+ 2 + S- 2 


x 10~ 22 

CoS (a) = Co+ 2 -f S- 2 


x 10~ 22 

PbS = Pb+ 2 + S" 2 


x 10- 26 

SnS = Sn+ 2 -f S- 2 


x 10- 26 

CdS = Cd+ 2 + S- 2 


x 10- 27 

Bi 2 S 3 = 2Bi +3 + 3S- 2 


x 10~ 70 

Cu 2 S = 2Cu+ + S~ 2 


x 10- 49 

Ag 2 S = 2Ag+ 4- S- 2 


x 10- 50 

CuS = Cu+ 2 + S- 2 


x 10~ 36 

Fe 2 S 3 = 2Fe+ 3 -f 3S~ 2 


x 10~ 88 

Hg 2 S = Hg+ 2 + S- 2 


x 10~ 45 

HgS = Hg^ 2 -f S- 2 


x 10- 50 

H 2 S is usually made by reacting acids with metallic sulfides, or heating 
sulfur with hydrocarbons. Some sulfides like A1 2 S 3 , Cr 2 S 3 , MgS, CaS, 
and BaS react with water to give H 2 S and insoluble metal hydrosulfides 
and hydroxides. Sulfur will dissolve in basic solutions to give various 
sulfides, depending upon the amount of sulfur added. Ions as S~ 2 , 
S 2 2 > S 3 2 > S 4 2 > S 5 2 > etc - are present. When acidified, these solutions yield 
H 2 S, S, and unfamiliar hydrides of sulfur as H 2 S 2 and H 2 S 3 . Cation 
group 3, 4, and 5 sulfides, except those of Co and Ni, dissolve readily in 
dilute HC1, those of Co, Ni, and group 2, except Hg+ 2 , require heating 
with stronger HC1, and aqua regia is needed to decompose HgS. The 
HC1 reactions give H 2 S, those with the oxidizing acid HNO 3 yield S; 
see list of reactions, Chapters 16 and 17. 


Sulfides are handled quantitatively by (a) oxidation to sulfate and 
precipitation and weighing as barium sulfate or (b) absorption of H 2 S in 
ZnAc 2 solution, followed by addition of a measured excess of I 2 and H + 
(to dissolve the ZnS formed and give S + I~), and titration of unreacted 
I 2 with standard thiosulfate. 

C. Sulfite Ion, SO 3 2 

Sulfurous acid is a diprotic acid made when SO 2 is dissolved in H 2 O. 
It is one of the weak acids, though dissociating more readily than most; 
KI = 1.25 X 10~ 2 . The sulfur oxidation state is (IV); intermediate 
among (II) in H 2 S; (0) in S; and (VI) in SO^ 2 , so sulfites may undergo 
reduction or oxidation. The partials show sulfurous acid to be a better 
oxidizing than reducing agent: 

3H 2 + S = H 2 S0 3 + 4H+ + 4e~ E = - 0.45 

H 2 + H 2 S0 3 = SOj 2 + 4H+ + 2e~ = - 0.20 

Sulfur dioxide gas is given off when alkaline sulfites are acidified and it 
may be identified by its pungent odor. H 2 SO 3 forms three kinds of salts: 
normal, MSO 3 , acid, MCHSQ^' and meta, MS 2 O 5 . Most sulfites and 
bisulfites except those of the alkali metals and NH+ are only slightly water 


Reactions (in order of decreasing [SO 3 2 ]) 


MgS0 3 6H 2 = Mg+ 2 + SCV + 6H 2 
ZnSO 3 -2H 2 O = Zn+ 2 + SO^ 2 + 2H 2 O 
BaS0 3 = Ba+ 2 + S0 3 
SrSO 3 = Sr+ 2 -f S<V 
Hg 2 S0 3 = Hg+ 2 + S0 3 


K (approximate) 

3.8 x 10- 3 

7.8 x 10~ 5 
8.3 x 10- 7 

3.9 x 10~ 8 
9 x 10- 28 

Qualitative testing for sulfite depends upon valence state changes, and 
release of SO 2 gas. Sulfites can be (a) reduced 

3Zn + 6H+ + H 2 S0 3 = H 2 S + 3H 2 O 
and the H 2 S tested as previously described ; (b) oxidized by strong oxidants 
5H 2 S0 3 = 4H+ + 5SOj 2 + 3H 2 O 


(r) precipitated and the solid changed in some characteristic manner 

PbAc 2 + SO 3 2 = PbS0 3>f + 2Ac- 
2PbS0 3 + 2 (air) = 

or (d) decomposed by heating with HC1 or H 2 SO 4 to give SO 2 , whose 
odor is unmistakable: 

2H+ + SO- 2 = (H 2 S0 3 ) = H 2 + S0 2 

In quantitative analysis, sulfites are titrated in acetic acid solution with 
standard 1% solution, using starch indicator or, alternately, SOg 2 is 
oxidized to SO 4 2 , precipitated, and weighed as BaSO 4 . 

D. Thiosulfate Ion, S 2 O~ 2 

Pure thiosulfuric acid is not isolated, but its salts are known and their 
stability likewise is not great. Acidification of thiosulfates results in 

2H+ + S 2 3 2 = (H 2 S 2 3 ) = S + H 2 + SO 2 

although in the presence of excess strong acid, H 2 S 2 O 3 appears to be 
stabilized and the reaction stops there for some hours. Most thiosulfates 
are water soluble and many dissolve in excess S 2 O 3 2 to give Werner 

The eventual precipitate of yellow-white, fine sulfur is one means of 
differentiating between S 2 O 3 2 and SO 3 2 . Thiosulfate is a reducing agent 
which gives tetrathionate or sulfite with mild oxidizing substances like I 2 , 
and S or SO 4 2 with stronger ones like MnO 4 . These reactions are /?H 
dependent : 

2S 2 3 2 = S 4 0- 2 + 2e~ E = - 0.17 

S 2 O 3 2 + 6OH- = 2SOg- 2 + 3H 2 O + 4e~ g = 0.58 

S 2 3 2 + 3H 2 = 2H 2 SO 3 + 2m + 4e~ E = - 0.40 

Na 2 S 2 O 3 is made by heating sulfur with sodium sulfite, and the penta- 
hydrate is the familiar "hypo" used in photography. There are many 
acids containing S, H, and O besides the four listed above. Some of 
these are sulfoxylic, H 2 SO 2 ; hyposulfurous, H 2 S 2 O 4 ; pyrosulfuric.U^O^ 
per oxymonosulf uric, H 2 SO 5 ; per oxydisulf uric, H 2 S 2 O 8 ; dithionic, H 2 S 2 O 6 ; 
irithionic, H 2 S 3 O 6 . 

Qualitative evidence of S 2 O 3 2 presence hinges on (a) destruction by acid 
to give SO 2 + S as already mentioned (b) bleaching of 1^ color as shown 
below (c) precipitation of white silver or lead thiosulfate, which on heating 
is converted to black silver or lead sulfide: 

2Ag+ + S 2 03 2 + H 2 = Ag 2 S 2 3jf + H 2 = Ag 2 S, + 2H^ + SO 4 2 


or (d) reaction in strong base solutions of cyanide to produce thiocyanate, 
which gives its characteristic red color with iron when Fe+ 3 + H + 
(HOOD!) are added. 

S 2 O 3 2 + CN- = SCN- + SO- 

Reactions (in order of decreasing [S 2 O3~ 2 ]) 

A' (approximate) 

SrS 2 O 3 -5H 2 O = Sr+ 2 + S 2 O 3 2 4- 5H 2 O 
BaS 2 3 = Ba+ a + S 2 O^ 2 
PbS 2 O 3 = Pb+ 2 + S 2 (V 
Hg 2 S 2 3 = Hg+ 2 + S 2 3 2 

8 x 10- 1 
9.5 x 10~ 5 
9 x 10~ 7 
1.8 x 10~ 35 

Thiosulfate solutions at approximately y?H 7 are titrated quantitatively 
using standard Ig~ solution and starch indicator. The reaction is: 

2S 2 Og 2 + I 3 = 31- + S 4 O~ 2 

In strong acid solution, S 2 O 3 2 is destroyed as noted above, and in strong 
basic solution, oxidation to sulfate can take place but the reaction is 
not stoichiometric : 

100H- + S^ 2 + 41- = 121" + 2S0^ 2 + 5H 2 O 

E. Carbonate Ion, COj 2 

When carbonates are acidified the reactions are 

+ 2H+ = HCOg + H+ = (H 2 C0 3 ) = H 2 O + C0 2 

Carbonic acid is an unstable, weak diprotic acid similar to sulfurous acid. 
The solubility of CO 2 in H 2 O is 1.50g/liter at 20 C and 1 atm, so the 
solution is 0.034 M. Most carbonates are water-insoluble but will decom- 
pose in acids stronger than H 2 CO 3 , as shown above. With the exception 
of the alkali metal carbonates, most others are decomposed by heat to the 
corresponding oxides. 

CaCO 3 - CaO + CO 2 

Suspended in water, CaCO 3 may dissolve if CO 2 is available, 

CaCO 3 + CO 2 + H 2 O = Ca+ 2 + 2HCOg- 
a reaction of importance in water works engineering. Slightly soluble 



calcium carbonate may thus be caused to change to the more soluble 
bicarbonate by cooling the solution to dissolve more CO 2 and vice versa. 
Alkali metal bicarbonates are less soluble than the corresponding car- 
bonates. See the discussion on H 2 CO 3 in Chapter 7. 

Carbonates and bicarbonates are identified by acid decomposition and 
absorption of CO 2 in lime water (Ca(OH) 2 ) or barytes water (Ba(OH) 2 ) 
to give a white alkaline earth carbonate precipitate: 

C0 2 + Ca+ 2 + 20H- = CaC0 3 

H 2 O 

To test for bicarbonate in the presence of carbonate, excess CaCl 2 
solution is added to precipitate carbonate, as above. The centrate from 
that is then treated with ammonia and half ihe (soluble) calcium bicar- 
bonate is convened to insoluble carbonate: 

Ca+ 2 + 2HCO3- + NH 3 = CaCO 3 + NH+ + HCOg- 

Mercuric chloride is also used to distinguish a bicarbonate solution from 
a carbonate; with the former it gives no reaction, with the latter it forms 
a brown basic carbonate. 


Reactions (in order of decreasing [CO 3 2 J) 


MgCO 3 3H 2 O = Mg+ 2 + COa~ 2 + 3H, 2 O 

1 x 10~ 5 

Ag 2 CO 3 = 2Ag f -f COa" 2 

8.2 x JO- 12 

NiCO 3 = Ni+ 2 -f CO," 2 

1.36 x 10~ 7 

CaC0 3 = Ca+ 2 + CO 3 2 (aragonite) 

6.9 x 10~ 9 

BaCO 3 = Ba^ 2 + CO^ 2 

1.6 x 10~ 9 

SrC0 3 = Sr+ 2 -f CO^ 2 

7 x 10~ 10 

CuCO 3 = Cu+ 2 + CO^ 2 

2.5 x 10~ 10 

ZnCO 3 = Zn+ 2 + CO^ 2 

2 x ID" 10 

MnCO 3 = Mn+ 2 + COg- 2 

8.8 x 10~ n 

FeCO 3 = Fe+ 2 + COf 2 

2.11 x 10- 11 

CdCO 3 = Cd 42 -f CO^ 2 

5.2 x 10- 12 

CoCO 3 = Co+ 2 + CO 3 2 

8 x 10~ 13 

PbCO 3 = Pb+ 2 + CO^ 2 

1.5 x JO- 13 

Alkali metal carbonates are determined quantitatively by titration with 
standard acid, a first end point noted which corresponds to HCOg~ 
formation, p\\ ^ 8.3, and a second end point corresponding to H 2 CO 3 , 
pH ^ 3.8. Alkaline earth carbonates are decomposed by heating to the 
oxide and CO 2 , loss in weight being used to compute carbonate content. 


Laboratory Study of the Group 1 Anions. Solutions of the anions 
contain lOmg of ion per milliliter and their preparation is listed in 
Appendix A 12. The student should pay attention to the features of 
analytical reactions which constitute positive tests for particular anions 
and also learn something of the interferences Lnd sensitivity of the tests. 
The same reactions will be used later to test unknowns. 

The instructor will comment on the notebook writeup for this work 
More important reactions are marked *. 

Test 22-1. Decomposition of NOg". (a) To 4 drops of test soln., add 
4 drops of H 2 O and a drop of dil. H 2 SO 4 . Note that bubbles of nitric oxide 
appear. What is the brown gas that forms above the soln? 

(/>) To 2 drops of NO^, add 8 drops of H 2 O, 2 drops of 0.1 A/sulfamic acid, 
a drop of 3 M H 2 SO 4 , and boil briefly. Cool. Neutralize with NH 4 OH. 

Test for NC>2~ via test 22-5 or another nitrite reaction. Explain. 

Test 22-2. Thiourea Test for NO%. To 3 drops of nitrite test soln,, 
add 3 drops of H 2 O, a few crystals of thiourea, and a drop of dil. HC1. Mix 
well. After bubbles of nitrogen have ceased to appear, add a drop of Fe+ 3 soln. 
To what is the red color due? Dilute the nitrite test soln. and rerun the test. 
Does the method have good sensitivity? How would one conduct this test if 
SCN" were originally present also in the sample? 

Test 22-3. Brown Ring Test for NO^. Add a drop of freshly prepd. 
FeSO 4 soln. to 2 drops of nitrite soln. and dil. with 6 drops of H 2 O. Keeping 
the tube inclined, carefully let 3-4 drops of dil. HAc or dil. H 2 SO 4 run down the 
incline and form a layer at the tube's bottom. Explain the formation of the 
dark ring at the interface. (Nitrates give a similar reaction if coned. H 2 SO 4 is 
used. Chromates, iodides, bromides, and ferrocyanides give colored products 
with iron and are removed by pptn. first with Ag 2 SO 4 .) 

Test 22-4. K 2 Na[Co(NO 2 )J from Nitrite. Mix 4 drops of nitrite test 
soln. with 3 drops of H 2 O, 4 drops of 2 M KC1, 4 drops of 0,5 M Co +2 , and 
2 drops of glacial HAc. Warm in a bath. The same yellow Werner compd. 
that appeared in the tests for Co +2 and K+ slowly ppts. Attachment to cobalt 
is via the N atoms, as - NO 2 groups. Give the equation. 

*Test 22-5. Diazo Test for Nitrite. Small amounts of nitrite may be 
detected by the diazo or Griess reaction. (See example 11, Chapter 12.) The 
interested student may wish to verify the sensitivity of the test. 

Dilute the nitrite test soln. 5-fold. Put 2 drops on a spot plate with 2 drops 
of sulfanilic acid and 2 drops of a-naphthylamine solns. A red color appears 
due to azo dye formation. Color density and formation time vary with the 
nitrite concn. Run a blank and compare. 

Test 22-6. Oxidation: NO% to NO^. To 4 drops of nitrite soln., add 
4 drops of H 2 O and 1 drop of dil. H 2 SO 4 . Dilute the KMnO 4 soln. on the 
shelf by three or four times and add this dropwise to the acidified nitrite, noting 


the bleaching of permanganate color, which indicates reduction of MnO 4 ~. This 
soln. could now be tested for nitrate (outlined under that ion) as proof that a 
nitrite was in the original sample. Write the equation and show via redox 
potential data that this reaction is expected. Could acidified chlorine water 
have been used as an oxidant? 

*Test 22-7. Testing for S~ 2 and H 2 S. Put 3 drops of sulfide test soln. 
in a tube, add 6 drops of H 2 O, and a drop of Pb+ 2 test soln. Black PbS is sufficient 
evidence with even the general anion unknown to report sulfide. 

By acidifying samples contg. considerable sulfide, H 2 S (POISON!) is detected 
by odor alone, even in the event other gases are also released. 

Several sulfides were mentioned in the descriptive section that normally 
react slowly with hot dil. HC1 but are caused to yield H 2 S if Zn metal were 
present. Prepare one and try the test. Write the equation. 

Test 22-8. Oxidation: S~ 2 to 5O 4 " 2 . With many oxidizing agents, S~ 2 
is oxidized to free sulfur but with some more drastic combinations like HC1 
+ HNO 3 , HNO 3 + Br 2 , or HNO 3 + KCIO 3 , sulfide is largely converted to 
sulfate, a change in oxidation state of eight units. 

To a mixt. of 2 drops of coned. HC1 and 3 of HNO 3 , add 2 drops of sulfide 
test soln. and heat carefully to slowly evap. the liquid. Cool. Add 30 drops of 
H 2 O and 10 drops of BaAc 2 soln. and mix. What is the slow-settling, fine, white 
ppt. ? Give equations. 

See special experiment 4, part 5 for another test of sulfur-contg. radicals. 

*Test 22 9. Production of SQ 2 and SO^ 2 from SO 3 2 . Put 5 drops of 
sulfite test soln. in the gas generating apparatus, Fig. 14-7, add 4 drops of H 2 O 
and 4 drops of 2 M HC1, warm the tube, and receive the distillate of gas in 1 ml 
of clear Ba(OH) 2 soln. A white ppt. of BaSO 3 , contaminated with a little 
BaCO 3 from CO 2 in the air, forms. 

To demonstrate that the ppt. is mainly BaSO 3 and not BaCO 3 , add enough 
HC1 to just dissolve it, then add 6-8 drops of satd. Br 2 water, and heat under 
the hood to expel bromine vapors. A white ppt. BaSO 4 (insol. in HC1) forms, 
showing SOa~ 2 was oxidized by Br 2 . Write equations. 

Test 22-10. SO^ 2 in the Presence of 5 2 O 3 2 . (a) Acids decompose 
thiosulfates to yield H 2 O -f S -f SO 2 so the presence of sulfur dioxide gas does 
not always mean that a sample under test contains sulfite. Sulfite may, however, 
be pptd. as the calcium salt since CaS 2 O 3 is sol., and this residue is then tested 
for SO 2 release. 

Mix 3 drops of each of the two test solns. with 4 drops of H 2 O. To this add 
5 drops of CaAc 2 soln., mix well, and cent. Wash the residue twice using a 
few drops of H 2 O each time and discard the washings. To the solid in the gas 
generating setup, add 3 drops of H 2 O and 6 of dil. HC1, quickly connect the 
delivery tube, warm, and collect the gas evolved in Ba(OH) 2 . Again demonstrate 
as in test 22-9 that SOi" 2 is in evidence. 

(b) Thiosulfates do not give the following test either, but sulfides and sulfites 
will bleach certain triphenylmethane dyes. Put 2 drops of malachite green 


on a spot plate and add dropwise a dil. neutral soln. of sulfite (the dye color 
changes with pti also, see chart of indicators, appendix). The reaction is given 
as example 10, Chapter 12. How could one remove sulfide in a mixture before 
this test? 

(c) Another method has been suggested by Lundin (see references p. 416). 
If a sample is suspected of contg. other ions such as S 2 C>3 2 , S~ 2 , and CO^" 2 , addn. 
of Sr+ 2 will precipitate only COj 2 and SO^ 2 . This residue is isolated by centri- 
fugation, then warmed with HC1, and the vapors tested with moist starch-iodate 
paper; SO 2 , but not CO 2 , reduces IO^ to I 2 , turning the paper blue. 

Test 22-11. Strong Oxidants and 5OJ 2 . Dilute 6 drops of SO^ 2 soln. 
to 1 ml and in another tube prep, a half ml of a mixt. of either MnOJ" or CrO^ 2 
solns. with an equal vol. of 2 M H 2 SO 4 . Add the oxidizing mixt. dropwise to 
the sulfite and account for the color change by writing the proper half reactions 
and E values from appendix A18. 

*Test 22-12. Decomposition of S 2 Og 2 . In the gas generator, acidify a 
few drops of thiosulfate test soln. with dil. HC1 and pass the gas into a clear soln. 
of barium hydroxide. What is the white ppt. in the receiver and opalescent 
suspension in the generator? Write the reactions. Does S~ 2 or SOs 2 interfere ? 

Test 22-13. Ag z S 2 O. 3 and its Decomposition. Mix 2 drops of thio- 
sulfate soln. with 3 drops of Ag+ test soln. on a spot plate, then observe the color 
changes. Explain with equations. (Excess S 2 C>3 2 will give the soluble complex, 
Ag(S 2 O 3 )~ 3 , but this is decomposable as above if boiled.) 

Test 22-14. S 2 O^ 2 in the Presence of SO^ 2 and 5~ 2 . Mix 4 drops of 
each of these test solns., then add 6 drops of H 2 O. Dropwise put in Zn+ 2 test 
soln. until pptn. is complete. Centrifuge and discard the white residue of ZnS. 
To the centrate, add Ca+ 2 test soln. until a new pptn. is complete, and discard 
the residue of calcium sulfite. (Ca and Sr sulfites are insoluble but their thio- 
sulfates are fairly soluble.) To the centrate add 4 drops of dil. HC1 and warm a 
little. The odor of SO 2 (CAUTION !) and ppt. of S proves S 2 Og 2 . Alternately, 
BaCl 2 soln. can be used to ppt. S 2 O3 2 and leave SO^ 2 in soln. after sulfide removal. 

' *Te$t 22-15. COg 2 in the Presence of SO~* and S 2 O^ 2 . In the gas 
generation apparatus, mix 4 drops of each of these 3 test solns., then add dil. 
KMnO 4 dropwise with stirring until a definite pink permanganate color is 
permanent above the ppt. of brown MnO 2 . This treatment oxidizes the sulfur- 
contg. radicals to sulfate. Now add 3 drops of dil. HC1, quickly insert the 
delivery tube, and bubble the gas into Ba(OH) 2 . The generator may be heated 
to increase the yield. A white residue in the barium soln. which is soluble in 
HC1 but not repptd. by Br 2 treatment (see test 22-9) is proof that the sample 
contd. carbonate. Give equations. 

Calcium hydroxide (lime water) may be used in place of barium hydroxide. 
The former is less sensitive, but the latter is more prone to interference by 
reaction with CO 2 in the air. A carbonate sample should give a very definite 
test, howe> -. not just a trace of turbidity. 


F. (Ortho)phosphate Ion, POj 3 

Of the several acids formed when P 4 O lo is dissolved in H 2 O, H 3 PO 4 
is the most important. A purer product is obtained by the action of 
nitric acid on white phosphorous or by reaction of sulfuric acid on a 
phosphate salt. Upon heating, phosphoric acid dehydrates to yield pyro- 
phosphoric, H 4 P 2 O 7 , and n-metaphosphoric, (HPO 3 ) n , acids in which n is 
2, 3, 4, 6, and possibly 1 . Solutions of pyro and metaphosphates are 
converted to orthophosphates on standing. Orthophosphate is the most 
important P v radical. Titration of H 3 PO 4 with NaOH can be followed 
by plotting pH versus volume of titrant (/?H curve) or by the use of 
selected indicators. Either shows sharp changes as the first and second 
H+ are neutralized. The third is too weak as an acid for accurate titration 
however. At the methyl orange end point, the salt formed is NaH 2 PO 4 , 
sodium dihydrogen orthophosphate, or primary sodium phosphate, and at 
the phenolphthalein point, is Na 2 HPO 4 , disodium hydrogen orthophosphate, 
or secondary sodium phosphate. Na 3 PO 4 is called tertiary or normal 
sodium orthophosphate. Its solutions are strongly basic due to hydrolysis. 
The phosphates of Na+, K+, and NHJ are soluble; most others including 
those of Li have low solubility. Pure H 3 PO 4 can be crystallized from 
solution by vacuum evaporation; its melting point is 42.3 C. It is a 
weak acid, ^ = 7.5 x 10~ 3 , K 2 = 6.2 x 10~ 8 . 

H 3 PO 3 + H 2 O = 2H+ + H 3 PO 4 + 2e~ E = 0.276 

HPO 3 2 + 30H- = 2H 2 + POj 3 + 2e~ Eg = 1.12 

Phosphorous and its compounds are derived from phosphate rocks: 
Ca 3 (PO 4 ) 2 + 3H 2 SO 4 = 2H 3 PO 4 + 3CaSO 4 

Ca 3 (P0 4 ) 2 + 5C + 6Si0 2 = 5CO 2 + 6CaSiO 3 + 4P 

Phosphorous chemistry is complex. The oxidation states are (111) 
as phosphine, PH 3 , (II) as hydrogen diphosphide, P 2 H 4 , (I) as sodium 
hypophosphite, Na 3 PO 2 , (III) as phosphorous trioxide, P 4 O 6 , (IV) as 
phosphorous tetroxide, P 2 O 4 , and (V) as phosphorous pentoxide, P 4 O 10 . 
(The tri- and pentoxides were named before the correct formulas were 
known and the names persist.) Phosphorous and a number of its com- 
pounds exhibit allotropy; P has three modifications and P 4 O 10 has four. 
The phosphorous acids are water soluble, but many of the salts are not. 
Phosphates are used as fertilizers, as emulsifiers in cheese and ice cream, 
in water treating chemicals to complex Ca +2 and Fe +3 , and in cleaning 
preparations, H 3 PO 4 and lime are used in raw sugar syrup purification. 


Organic phosphates have also become important: some are used in 
insecticides (parathion, schradam, systox), while some fluoro organic 
phosphates form a new series of unusually deadly "nerve" gases for 
warfare. Tricresyl phosphate (T.C.P.) is a gasoline additive. These 
phosphates are not water soluble but can be made to yield PO 4 3 tests 
after hydrolysis by hot alkalis. 


Reactions (in order of decreasing [PO 4 - 3 ]) 


Mg(NH 4 )P0 4 = Mg+ 2 + NH 4 f + POif 3 
Mn 3 (PO 4 ) 2 = 3Mn+ 2 + 2PO 4 3 
Sr 3 (P0 4 ) 2 = 3Sr+ 2 + 2PO 4 3 
Ca 3 (P0 4 ) 2 = 3Ca+ 2 + 2PO4- 3 
Ba 3 (P0 4 ) 2 = 3Ba+ 2 + 2P(V 
Fe(P0 4 ) = Fe^ 3 -f PO^ 3 
Pb 3 (P0 4 ) 2 = 3Pb+ 2 -h 2POJ 3 

2.5 X 10- 13 

~ io- 22 

1 x 10~ 31 
1.3 x 10~ 32 
6 x IO- 39 
1.5 x 10~ 18 
1 x IO- 54 

Large quantities of phosphate are quantitatively determined by precipita- 
tion as MgNH 4 PO 4 , ignition to Mg 2 P 2 O 7 , and weighing in that form. 
Small quantities of PO^ 3 are handled colorimetrically. See the molybdate- 
benzidine test. 

G. Sulfate Ion, SO 4 2 

Since the oxidation state of S here is (VI), the highest it can attain, 
sulfate may function only as an oxidizing agent: 

H 2 O + H 2 SO 3 = 4H+ + SO 4 2 + 2e~ E = - 0.20 

With hot, concentrated suljuric acid, reducing agents liberate sulfur 

2H 2 SO 4 + Zn = Zn+ 2 + SO 4 2 + H 2 O + SO 2 

Pure H 2 SO 4 is an oily liquid only about 2% dissociated. It forms a 
number of hydrates, as H 2 SO 4 -H 2 O, etc., which are quite stable and 
account for the speed and efficiency with which the acid dehydrates 
materials. In dilute water solutions, the first ionization is considered 
100% complete and the second is appreciable. SO 3 dissolved in H 2 SO 4 
is called fuming sulfuric acid or oleum, and when hot, the mixture is a 
powerful oxidizer. H 2 SO 4 is made by absorbing SO 3 in H 2 SO 4 giving 
H 2 S 2 O 7 , which upon dilution gives more H 2 SO 4 . The SO 3 is prepared 


from oxidation of SO 2 , which in turn is made by burning sulfur in air. 
H 2 SO 4 costs only a few cents a pound and is industrially very important. 
It is used to prepare other chemicals like phosphate fertilizers, for pickling 
metals prior to surface treatment, for absorbing unsaturates and other 
impurities from alkanes in petroleum refining, in neutralizing basic solu- 
tions for the production of synthetic fibers, etc. Natural sulfates like 
gypsum, CaSO 4 -2H 2 O, occur in many minerals and waters and are used 
as a source of sulfate and in making plaster and cement. With the 
exception of the sulfates in the following table, the other common ones 
are fairly soluble. 


Reactions (in order of decreasing [SOj 2 ]) 


Ag 2 S0 4 = 2Ag+ 4- S0 4 2 
CaSO 4 2H 2 O = Ca+ 2 + SO^ 2 + 2H 2 O 
Hg 2 S0 4 = H& 1 2 + S0 4 2 
SrS0 4 = Sr+ 2 + S0 4 2 
PbS0 4 = Pb t2 + S0 4 2 
BaSO 4 = Ba+ 2 + SO 4 2 

1.24 x 10~ 5 
2.4 x 10~ 5 
~ 1 x 10- 6 
7.6 x 10~ 7 
1.3 x 10- 8 
1.5 X 10- 9 

A number of titrations have been proposed for the quantitative analysis 
of sulfate, using BaCl 2 , alcohol-water solutions, and indicators like 
alizarin, but none of these rival the accuracy of the old gravimetric 
method in which BaSO 4 is precipitated, ignited at 800 C, and weighed. 
The same precipitate is used in qualitative testing to indicate sulfate; it 
is insoluble in HC1. 

H. Chromate, CrO 4 2 , and Bichromate, Cr 2 O~ 2 , Ions 

Bichromate ion is formed in acid solution, and chromate ion in basic 
solution. Equilibrium is established between the two as 

H 2 = 2Cr0 4 2 + 2H+ 

]*[H+] 2 /[Cr 2 0- 2 ] = 2.4 x 10~ 15 

From this it is seen that in an acidic solution, say pH 2, the ratio 
[CrO 4 2 ] 2 /[Cr 2 O7 2 ] = 2.4 x 10~ n and at apH of 12, the ratio is 2.4 x 10 9 . 
Further condensation to progressively darker Cr 4 Oj 2 and Cr 5 O^ 3 2 is 
known but is of minor importance. 


Since chromates and dichromates are good oxidants they cannot be 
present in solution with good reductants: 

2Cr+ 3 + 7H 2 = Cr 2 07 2 + 14H+ + tor E = - 1.33 

All chromates and dichromates are colored. Aqueous solutions of 
CrO 4 2 are yellow and of Cr 2 O7 2 are orange. Most dichromates are water 
soluble. The insoluble chromates in Table 22-7 dissolve in strong acid. The 
chromates of Hg+ 2 , Fe+ 2 , Mn+ 2 , and Sr+ 2 are a little more water, soluble. 


Reactions (in order of decreasing [CrO 4 2 ]) 

CaCrO 4 = Ca +2 + CrO 4 2 
SrCrO 4 = Sr+ 2 -f CrO;f 2 
Hg 2 Cr0 4 - Hg 2 + 2 + Cr0 4 2 
Ag 2 CrO 4 = 2Ag+ + Cr0 4 2 
BaCrO 4 = Ba+ 2 -f- CrO 4 2 

PbCrO 4 = Pb+ 2 -f CrO 4 2 


7.1 x 10~ 4 

3.6 x 10~ 5 

2 X 10~ 9 

1.9 X 10~ 12 

8.5 x 10- 11 

2.0 x 10~ 16 

Chromates are used in tanning, in yellow paints, and as corrosion 
preventatives in water treatment. 

Chromates and dichromates are determined quantitatively by reaction 
with excess I" to give a stoichiometric amount of I~, which is titrated 
with S 2 O3 2 using starch indicator. For small amounts of chromium in 
the (VI) state, diphenylcarbazide is an excellent colorimetric reagent. 

See Chapter 17 for further Cr chemistry. 

I. Borate Ions, BOg-, BOj 3 , B 4 C>7 2 

These ions contain boron at an oxidation state of (III) and depending 
on temperature, pH and concentration, one or another predominates in 
solution. Acidification of borates gives H 3 BO 3 , orthoboric, or boracic acid. 
Heating converts it at 100 C to HBO 2 , metaboric acid, then to H 2 B 4 O 7 , 
tetra- or pyroboric acid, at 140 C, and finally to B 2 O 3 , boric oxide, the 
common anhydride. Important reactions are, 

B + 4OH- = BO 2 + 2H 2 O (only in strong base) 
B 2 O 3 + 2OH~ = 2BO2 + H 2 O (only in strong base) 
H 3 BO 3 + OH- = 2H 2 O + BOg = HBO 2 + OH~ + H 2 O 
BOg 3 + 2H 2 O = 2OH- + H 2 BO^ 

2H 3 B0 3 + 2B0 2 = B^ 2 -f- 3H 2 O = HBO 2 + 2BO 2 + 2H 2 O 
5H 2 2H 3 BO 3 + Ba(H 2 BO 3 ) 2 (or Ba(BO 2 ) 2 -2H 2 O) 


H 3 BO 3 is a waxy solid, slightly soluble in H 2 O at room temperature, giving 
reactions of a very weak monoprotic acid. It is used as an antiseptic. 
Borax, sodium tetraborate decahydrate, Na 2 B 4 O 7 - 10H 2 O, is the most 
common boron derivative. It is used in the manufacture of cleaners, 
water softening chemicals, soaps, glass, embalming fluid, etc. Alkali 
metal borates are soluble but others are usually only slightly soluble in 
water, though soluble in strong acids since boric acid is very weak. 

Recent work in boron chemistry has been along two dissimilar lines: 
"super" abrasives and fuels. In 1957 the preparation of "Borazon," 
cubic boron nitride, was announced by General Electric. The cubic form 
was not known previously, and the structure was proved by X-ray diffrac- 
tion. It is harder than diamond and much superior in oxidation resistance. 
It will be used in cutting tools. Boron hydrides, diborane B 2 H 6 , pentaborane 
B 6 H 9 , and decaborane B 10 H 14 , are being used as rocket fuels. When 
burned with liquid O 2 , 40 % more thrust is developed than when kerosene 
is burned under the same conditions. These hydrides are hypergallic in 
air, are toxic, and cost up to several thousand dollars per pound at this 
time. Some borate esters are added to gasoline to improve antiknock 
quality for motor car use. 

One qualitative test for borates is acidification and treatment with an 
alcohol to form a volatile borate ester. When this is vaporized and 
burned, it imparts a green color to the flame : 

B 4 O~ 2 + 5H 2 O + H 2 SO 4 = H 3 BO 3 + SO^ 2 

H 3 B0 3 + 3C 2 H 5 OH = 3H 2 + B(OC^i b \ 
2B(OC 2 H b \ + 50 2 = 5H 2 + 4CO 2 + B 2 O 3 

Another test is with turmeric paper, which gives a brown color with 
borates + HC1, and after drying and rewetting with NaOH, the color 
turns dark green. Oxidizing agents interfere because they bleach the 
reagent. These are tests 22-23 and 22-25. 


Reactions (in order of decreasing [BO 2 ]) 

LiBOo = Li+ + BO. 

Ca(BO 2 ) 2 -6H 2 O = Ca+ 2 + 280^ + 6H 2 O 

AgB0 2 = Ag+ + B0 2 

Mn(BO 2 ) 2 2H 2 O = Mn+ 2 + 2BOJ + 2H 2 O 

K (approximate) 

4.5 x 10- 1 
1.2 x 10~ 4 

3.6 x 10- 3 
3.9 x 10~ 7 


Borax can be titrated with standard acid using methyl orange indicator. 
It acts like other salts of weak acids. Small amounts of borates are 
quantitatively found using colorimetric reagents. (See Chapter 12 on 
organic reagents for borates.) Borates in moderate amounts may also 
be determined by reaction with a poly alcoho 1 like glycerol or mannitol 
and titration of H+ liberated using standard base : 

OH OH O/ \> 

H 3 B0 3 + C C = C --- C + 2H 2 + H+ 

J. Fluoride Ion, F~ 

HF is not a reducing acid as the other halogen acids may be, since no 
chemical oxidant is powerful enough to liberate F 2 from fluorides. In 
the vapor state it forms polymers like (HF) n , where n has an average value 
of 6 at room temperature. With H 2 O, it forms an azeotrope boiling at 
120 C, which is about a 12 M solution. Both the vapor and solutions 
produce slow healing burns. HF will dissolve glass and silicate minerals, 
giving H 2 SiF 6 and so is handled in lead lined equipment and dispensed in 
wax or plastic bottles. Fluoride ion forms strong complexes such as 
AlFg 3 , BeF^ 2 , SnF~ 2 , FeF~ 3 , and ZrF^ 3 . 

Hydrofluoric acid is used as an etchant and in the pickling of alloys 
like stainless steel after heat treating. Sodium fluoride is used in fluorida- 
tion of water, as F~ at a concentration of 0.8-1.6 ppm has been found 
to be beneficial for children's teeth; concentrations much greater result 
in mottled enamel, however. Teflon is a commercial polymer 
(CF 2 CF 2 ) which is unusually inert and high melting. It is used as a 
gasket material, etc., in the most corrosive atmospheres. 

OF 2 , NF 3 , and F 2 have been tried as rocket propel lant oxidizers. 
Fluorine burning with kerosene, hydrazine, or liquid hydrogen gives 
higher energies than liquid ozone and the same fuels. Liquid F 2 is trans- 
ported in nickle alloy containers kept cool with jackets of liquid nitrogen. 

The H F bond has probably the highest heat of formation of any 
measured; even solid fluorine and liquid hydrogen at 252.5 C explode 
when mixed. HF is the most associated substance known in aqueous 
solution. The equilibria are: 

HF = H+ + F- #! = 6.71 x 10- 4 

HF + F- = HF 2 K 2 = 5.1 

and because of this, HF is a weak acid, in sharp contrast to other halogen 


acids of the type HX. Oxygen fluorides are OF 2 and O 2 F 2 ; the former, 
more common, is made by bubbling fluorine gas into 0.5 M NaOH. 

Fluorine is one of the strongest oxidants known and is prepared only 
by electrolysis : 

2F - = F 2 + 2e~ E = - 2.8 

The fluorides of silver and the alkali metals are soluble, but most 
others are not, though they dissolve in strong acids. 


Reactions (in order of decreasing [F~]) 


BaF 2 = Ba+ 2 + 2F~ 
MgF 2 = Mg- 2 + 2F~ 
PbF 2 = Pb+ 2 + 2F~ 
SrF 2 = Sr+ 2 + 2F~ 
CaF 2 = Ca+ 2 + 2F~ 

2.4 x 10~ 5 
8 x 10~ 8 
4 x 10- 8 
7.9 x 10- 10 
1.7 x 10- 10 

Larger quantities of fluoride can be segregated for testing by distillation 
from sulfuric acid in glass, as H 2 SiF 6 , and titrated quantitatively with 
thorium nitrate using zirconyMizarin indicator. Small amounts are 
done colorimetrically using the same reagent. (See special experiment 3.) 
A1+ 3 is an interference, since it is capable of complexing F~ and preventing 
its reaction. PO 4 3 and C 2 O 4 2 also interfere by bleaching the reagent to a 
yellow color just as F~ does. All these hindrances are obviated by 
distilling, as mentioned above, and testing the distillate for F~. 

K. Oxalate Ion, C 2 O 4 2 

Oxalic acid is readily crystallized from aqueous solution as the dihydrate, 
H 2 C 2 O 4 2H 2 O. Strong heating decomposes it to CO, CO 2 , and H 2 O. 
It is one of the stronger weak acids, A^ = 3.8 x 10~ 2 , K 2 = 5.0 x 10~ 5 . 
Oxalates in general are not too water soluble but will dissolve in strong 
acids because H 2 C 2 O 4 is displaced. Some oxalates dissolve in an excess 
of C 2 O 4 2 because of complex formation. 

Qualitative testing of oxalate is best done by showing that Ca+ 2 precipi- 
tates white calcium oxalate which is soluble in dilute sulfuric acid (a little 
CaSO 4 may precipitate) and the oxalic acid produced decolorizes potassium 
permanganate solution: 

CaC 2 O 4 + 2H+ + SO 4 2 = Ca+ 2 + SO 4 2 + H 2 C 2 O 4 
5H 2 C 2 O 4 + 2MnOi- + 6H+ = 8H 2O + 10CO 2 + 2Mn+ 2 



Reactions (in order of decreasing [C 2 O 4 2 ]) 

MgC 2 4 = Mg+ 2 + C 2 4 2 

SrC 2 4 H 2 = Sr+ 2 + C 2 O 4 2 + H 2 O 

BaC 2 4 2H 2 = Ba+ 2 + C 2 O 4 2 -f 2H 2 O 

CdC 2 O 4 = Cd+ 2 -f C 2 O 4 2 

CaC 2 4 H 2 = Ca+ 2 + C 2 O 4 2 -f H 2 O 

PbC 2 O 4 = Pb+ 2 + C 2 O 4 2 

8.6 x 10- 5 
5.61 x 10- 8 
1.5 x 10~ 8 
1.5 x 10- 8 
1.3 x 10~ 9 
8.3 x 10~ 12 

Sodium oxalate is synthesized commercially by heating CO and NaOH 
in an autoclave; sodium formate is an intermediate. Oxalates are used in 
tanning, ir certain cleaning preparations, and in dye manufacture. 

Quantitatively, oxalates are precipitated as CaC 2 O 4 and ignited to and 
weighed as CaO, or the calcium oxalate is dissolved in H 2 SO 4 and the 
oxalic acid titrated, as shown above, with standard permanganate. 

Laboratory Study of the Group 2 Anions 

,/*Test 22-16. Ammonium Molybdophosphate. To 4 drops of PO 4 3 
test soln. add 4 drops of H 2 O and 4 of coned. HNO 3 . Shake. Then add 8 drops 
of ammonium molybdate, (NH 4 ) 6 Mo 7 O 24 , and heat the tube in a water bath for 
a few min. A slow-forming yellow ppt. is the ammonium salt of a heteropoly 
ion. (See Chapter 4.) This material has a variable composition dependent 
upon pptn. conditions, but the ratio of PO 4 to MoO 3 is 1/12 and the accepted 
formula is (NH 4 ) 3 [P(Mo 3 O 10 ) 4 ]-2HNO 3 -2H 2 O. The complex radical is some- 
times given as (PO 4 - 12MoO 3 )- 3 . The ammonium molybdate reagent contains 
some NH 4 NO 3 , furnishing the common nitrate ion to decrease the solub. of 
the ppt. A known should be run since the reagent deteriorates slowly upon 
standing. Arsenates, which are not considered here, also give the test but may be 
treated first with CH 3 CSNH 2 and NH 4 I in acid soln. to ppt. As 2 S 3 . 

Test 22-17. Reactions of Ag+, Ba+ 2 , Ca+\ Mg+ 2 , and Fe+* with PO^ 3 . 

(a) Put 4 drops of PO 4 3 soln. in a test tube, add a drop of dil. NH 4 OH and 
3 drops of Ba+ 2 test soln. Repeat using Ca+ 2 and Mg+ 2 in place of Ba+ 2 . 
What are the ppts.? Omitting the ammonia, which would complex silver, 
show that yellow silver phosphate, Ag^O^ is also pptd. Try the solub. of 
these residues in 6 M HC1. Write the equations. 

(b) To 6 drops of Fe+ 3 soln. add a drop of 3 M H 3 PO 4 . Note the colors of 
the soln. and the ferric phosphate ppt. Now, with stirring, add more H 3 PO 4 
dropwise and describe the change. Soluble complex acids of the type Fe(HPO 4 ) 
(H 2 PO 4 ) and Fe(H 2 PO 4 ) 3 are present. Try the SCN~ test for Fe+ 3 and explain. 


How does PC>4 3 hinder the analysis of Fe+ 3 in a general unknown? How 
could the phosphate complexes be decomposed so that one could get a good 
iron test? Try a method you propose and report its result. 

Test 22-18. Molybdophosphate-Benzidtne Oxidation-Reduction. 

(a) A sensitive variation of test 22-16 is treatment of the molybdophosphate 
with benzidine. The organic reagent is oxidized to benzidine blue (example 8, 
Chapter 12), and the heteropoly ion is reduced to molybdenum blue (test 20-6), 
the result being a dark blue color even with quantities of phosphate too small to 
give a ppt. with ammonium molybdate. Arsenates and free molybdic acid 
do not give the test. 

Put a drop of dil. phosphate soln. on a spot plate and add a drop of ammonium 
molybdate and a drop of benzidine. In the next depression on the plate, put 
2 drops of dil. NH 4 OH. With a micro stirring rod, transfer small quantities of 
ammonium hydroxide into the other soln. A dark blue coloration is a positive 

(b) Modification is possible using spot technique on filter paper, or other 
reducing agents like SnCl 2 in HC1. Try a variation and report the results. The 
concn. limit using Sn 11 on the molybdophosphate is about 0.1 ppm PO^ 3 . 

*Test 22-19. BaSO^ and its Solubility. Add 6 drops of H 2 O to 4 
drops of SO^ 2 soln., then add 3 drops of Ba+ 2 test soln., and shake. Centri- 
fuge and test the solub. of the residue in hot 6 M HC1. How does this distinguish 
sulfate among ppts. of the other group 2 barium and calcium salts? The 
method outlined in part 2(b) of special experiment 4 illustrates one method of 
confirming the ppt's identity. 

Test 22-20. Phenolphthalein- Barium Carbonate Test for SO^ 2 . 

In a crucible, evap. 6 drops of an Na 2 SO 4 soln. with 50 mg of solid BaCO 3 by 
heating on a water bath. Allow it to cool when evapd., then add a drop of H 2 O 
and a drop of 1 % phenolphthalein. A pink coloration shows that the pH has 
increased due to formation and subsequent hydrolysis of a sol. carbonate 
produced from a transposition reaction. Write the equations. The concn. 
limit is about 150 ppm SO^ 2 . 

*Test 22-21. BaCrO^; Oxidation to CrO 6 . To 4 drops of chromate 
soln., add 2 drops of 2 M NaAc, 6 drops of H 2 O and, 5 drops of Ba+ 2 test soln. 
centrifuge and wash the residue, rejecting the washings. Show that the BaCrO 4 
is sol. in a few drops of dil. HNO 3 . To this orange soln., now contg. dichromate 
ion predominantly, add a ml of H 2 O and 5 drops of -butyl alcohol. Put in 
5 drops of 3 % H 2 O 2 and shake cautiously. The appearance of a blue color in 
the alcohol layer is due to peroxychromic anhydride, CrO 5 . 

Test 22-22. CV 2 Of 2 as an Oxidizing Agent. To 6 drops of 0.25 M 
SnCli" 2 soln., add 2 drops of 6 M H 2 SO 4 and 4 drops of H 2 O. Now add Cr 2 O7 2 
or CrOi~ 2 soln. dropwise and note the evidence of reaction. Write the equations 
and relate to their oxidation potentials. To what is the green color due? 


*Test 22-23. Ethyl Borate. Mix 5 drops of B 4 O^ 2 test soln. with 8 drops 
of ethyl alcohol in a crucible. Add 3 drops of con. H 2 SO 4 , stir, then warm 
gently. Ignite the ethyl borate vapors and note the green flame, characteristic 
of B 2 O 3 . Do not inhale the fumes! The color is best observed in a semidark 
room. Methyl alcohol is a slightly more sensitive reagent. What are the 
reactions? Why must Cu+ 2 and Ba+ 2 be removed, if present, prior to making 
this test? 

Test 22-24. Precipitation and Solution of Barium Metaborate. 

Calcium and barium metaborates are sol. in excess reagent, strong acids, or 
ammonium solns. Demonstrate this as follows: put 1 ml of B 4 O^ 2 soln. in a 
tube, add Ba+ 2 until a heavy ppt. is present, then divide it among 3 tubes. To 
one add 5 drops of 6 M HC1, to another 10 drops of Ba +2 soln., and to the last 
add 75 mg of solid NH 4 C1. Stir each and note the results. 

Test 22-25. Turmeric Test for Borates. On a spot plate make a mixt. 
of a few drops of borate soln. and a drop of 6 M HC1. Put 2 drops of this on a 
piece of turr *n'c paper and dry it by waving over a low flame. Note the color. 
Now add a chop of 0.25 M NaOH to the spot and place another NaOH drop 
on another part of the paper. Compare the test spot color with the latter 
"blank." What substances interfere if not removed first? The reaction will 
detect as little as 1 ppm boron. 

Test 22-26. Organic Reagents for Borates. A number of excellent 
organic reagents are capable of detecting low concns. of boron (as borates) 
without interference by moderate quantities of other common ions except 
strong oxidizing agents. (See example 3, Chapter 12, and part 1 of special 
experiment 8.) 

(a) Put a drop of H 2 O in one depression of a spot plate, a drop of 1 : 10 borate 
soln. in another, and a drop of undil. borate soln. in a third. To each, in order 
given, add a drop of coned. HC1 and 2 drops of coned. H 2 SO 4 . Wait a minute 
for any NO^ which might be present to be decomposed, since it bleaches the 
reagent. Now add a drop of carmine soln. (appendix A15) to each. Give the 
plate a circular swirling motion to mix each spot, then note the color changes. 

(b) Repeat using quinalizarin soln. instead of carmine. 
Both methods are used in quant, colorimetry for boron. 

*Test 22-27. Alizarin-Zirconyl Test for F~. This test will detect 
0.1 ppm F~ and is used quantitatively for fluoride in the range 0.1-3 ppm. 
Larger concns. give a uniform color. Although this range is small, it is the 
important one in water works chemistry. See example 5, Chapter 12, p. 194, 
and special experiment 3. 

(a) Put I drop of F~ soln. in a tube and add 29 drops of H 2 O; in a second 
tube put 2 drops of F~ and 28 drops of H 2 O; in a third put 8 drops of F~ and 
22 drops of H 2 O; in a fourth, put 2 ml of tap water; and in a last, put 2 ml of 
distilled water. To each add 2 drops of alizarin-zirconyl reagent (appendix A 1 5) 
and shake. After 1 5-30 min observe the colors by looking through the tubes 
vertically as they are held before a white background. Note the color gradations. 


(b) Oxalate and phosphate were mentioned in the fluoride discussion as ions 
which give false tests with the fluoride reagent. Try one or both and compare 
with previous colors. Suggest a way to remove either or both and give your 
plan a trial. Describe the results. 

Test 22-28. Precipitation of CaF 2 . To 4 drops of F~ soln., add 4 drops 
of H 2 O, then 4 drops of Ca+ 2 . Centrifuge. To a part of the centrate try 
test 22-27. Make a 1 : 5 diln. with some of the remaining centrate and repeat 
test 22-27. What has been demonstrated? 

Test 22-29. Etching Glass. With large quantities of F~ in a sample, 
this cumbersome method is sometimes used but it has been largely replaced by 
test 22-27 or other color tests of that principle. 

Mix 500 mg of a solid fluoride with 2 ml of 6 M H 2 SO 4 in a small lead dish 
in the hood. Cover it with a waxed glass plate, through which coating has been 
scratched some identifying marks. Warm the dish intermittently. After 
2 hours remove the wax by warming and wiping, then examine the clean glass 
surface for evidence of HF attack. 

*Test 22-30. Precipitation of CaC 2 O 4 ; Oxidation with MnO^. To 

6 drops of oxalate soln., add 4 drops of 6 M HAc, then calcium test soln. until 
pptn. is complete. Centrifuge, save the residue. To this add 6 drops of H 2 O 
and 2 drops of 6 M H 2 SO 4 and stir to dissolve the CaC 2 O 4 . Warm the tube, 
then add 0.01 M KMnO 4 drop wise and observe the decolorization denoting 
oxidation of oxalate. What gas is given off? 

Test 22-31. Thermal Decomposition of H 2 C 2 O 4 . In a gas generating 
tube, i/i the hood, heat a mixt. of 4 drops of oxalate soln. and 4 drops of coned. 
H 2 SO 4 , and bubble the distillate into Ca(OH) 2 or Ba(OH) 2 . What gases are 
given off? Which one reacts to produce the white precipitate? How could 
one analyze for the other ? 

Test 22-32. Resorcinol Test for C 2 O^ 2 . The reaction for this test is 
described as example 14, Chapter 12. To 4 drops of oxalate soln. in a small 
beaker, add a drop of 6 M H 2 SO 4 and 4 drops of H 2 O, then cautiously sprinkle 
in a little magnesium powder. When the metal has dissolved, pour the soln. 
into a test tube, add 2 drops of resorcinol reagent, mix, then carefully allow 
6-7 drops of coned. H 2 SO 4 to run down the side and form a layer at the bottom. 
A blue to red layer indicates glycolic acid, which in turn means a positive test 
for oxalate. The concn. limit is about 20 ppm C 2 O 4 2 . Other common organic 
acids such as acetic and citric do not give the test. 

L. Chloride Ion, Cl~ 

This ion is a very weak reductant: 

2C1- = C1 2 + 2e~ E = - 1.36 


Oxidation to give chlorine gas occurs only with such potent oxidizers 
as F 2 , MnO^, BiO^, and S 2 Og 2 . HC1 is a pungent gas made by direct 
combination of the elements or by heating NaCl with concentrated 
sulfuric acid : 

H 2 + C/ 2 = 2HCI 

NaCl + H 2 S0 4 = NaHS0 4 + HCI 

HC1 gas is not very soluble in nonionizing solvents but is quite soluble 
in water and yields hydrochloric acid, which in technical grade is known as 
muriatic acid. Infrared absorption study of the aqueous solutions gives 
no bands characteristic of HCI, showing that complete hydration has 
taken place. Oxides of chlorine are C1 2 O, C1O(?), C1O 2 , and C1 2 O 7 in 
which the oxidation states of Cl are respectively (I), (II), (IV), and (VII). 
The most familiar valence, however, is (1) as typified by chloride ion, 
Cl". It is detected, after oxidation or previous precipitation of inter- 
fering ions (such as Br~, I~, SCN~, and S~ 2 ) by reaction with Ag^, with 
which it forms a curdy white precipitate, soluble in NH 4 OH. The 
Ag(NH 3 ) so formed is reprecipitated as silver chloride with HNO 3 . 
See Chapter 15 for reactions. 

With the exception of SbOCl and the chlorides in Table 22-11, most 
are water soluble and all show increased solubility in solutions containing 
excess Cl~ due to complex ion formation. The chlorides of Pb f2 , Hg+ 2 , 
and Cd+ 2 are weak electrolytes. 


Reactions (in order of decreasing [Cl~]) 


PbCI 2 = Pb+ 2 + 2C1- 
CuCl = Cu+ + Cl~ 
BiOCI = BiO+ + Cl- 
AgCI = Ag+ + Cl- 
Hg 2 Cl 2 = Hg+ 2 + 2C1" 

1.6 X 10~ 5 
3.2 x 10- 7 
7 x 10~ 9 
2.8 x 10- 10 
1.1 x 10- 18 

There are many uses for chlorides such as NaCl and CaCl 2 and these 
are mentioned with discussions of the cations. 

Quantitatively, chloride is determined by titration with standard 
AgNO 3 or by precipitation and weighing as AgCI. 

M. Bromide Ion, Br~ 

HBr is a colorless gas and like HCI is not very soluble in nonassociating 


solvents but dissolves in H 2 O to give a strong acid. Anhydrous liquid 
HBr is no more ionized than is H 2 O. Bromides are similar to chlorides 
in general properties; the bromides of the group 1 cations are less soluble, 
however, and Br~ is more readily oxidized than Cl~. Bromides occur in 
natural brines and Br 2 is obtained by C1 2 displacement commercially: 

C1 2 + 2Br~ = Br 2 + 2C1~ 

Other, more expensive oxidizing agents one could employ here include 
HC1O, Cr 2 Of 2 , MnO^, PbO 2 , MnO 2 , etc., in acid solution. The bromide- 
bromine partial is 

2Bi- = Br 2(ftq) + 2e~ E = - 1.09 

HBr is not made from direct combination of elements because the low 
heat of formation (4.8 kcal/mole) precludes a good yield. It is best 
prepared by distillation from a mixture of bromide salt and H 3 PO 4 , the 
latter used for its high boiling point and nonoxidizing character. AgBr 
is used in some photographic emulsions and a few bromides are used in 
medicine. HBr finds little use. 

Qualitatively, bromide is detected by precipitation as light yellow AgBr, 
which is soluble only in fairly concentrated NH 4 OH or by oxidation to 
Br 2 , which is extracted in CC1 4 , giving a brown solution. 


Reactions (in order of decreasing [Br~]) 


Pbfcr 2 = Pb+ 2 + 2Br~ 
CuBr = Cu f + Br~ 
AgBr = Ag+ + Br~ 
Hg 2 Br 2 = Hg+ 2 + 2Br- 

4.6 X 10- 6 
6 x 10~ 9 
5 x 10~ 13 
1.3 x 10~ 22 

Quantitatively, bromide may be determined by precipitation with Ag+ 
and weighing the AgBr, or by titration with standard Ag f . 

N. Iodide Ion, I 

Iodide is a reducing agent in acid solution, where, depending upon the 
oxidizer's strength, it may give iodine or iodate. These partials are 
important in understanding the selective oxidation of I~ in the presence 
of other oxidizable substances: 

21- = I 2 + 2e~ E = - 0.54 

I- + 3H 2 O = IOr + 6H+ + tor E = - 1.09 


For instance Table A 18 shows that HNO 2 is capable of oxidizing I~ but 
not Br~ or Cl~. Reaction equations can be formulated accordingly: 

2H+ + 2HNO 2 + 21- = I 2 + 2NO + 2H 2 O 

The I 2 can be extracted with CC1 4 and the aquer us layer tested for other 
ions like Br~. 

Nonassociating solvents such as CHC1 3 and CC1 4 give purple solutions 
with iodine, others such as H 2 O and alcohols give brown solutions 
containing triiodide ion, Ij. I 2 is quite soluble in aqueous I~ solution 
for that reason. Other iodides also form complexes with excess I~ 
which are soluble: 

Hg+ 2 + 41- = HgI 2R + 21- = HgI 4 - 2 

With cupric ion, an oxidation-reduction takes place and iodine and 
cuprous iodide precipitate. If sulfite is present, the iodine is reduced to 
iodide again, leaving only the Cul precipitate. 

SOj 2 + 2OH- + 2Cu+ + 21- = 2CuI FF + SOj 2 + H 2 O 

Iodine reacts slowly with water to give iodide and hypoiodous acid; 
the latter' s decomposition is aided by light thus promoting I 2 hydrolysis. 

I 2 + H 2 = I- + HIO + ff 
2HIO = 21- + 2H+ + O 2 

Pure hydriodic acid is obtained by heating an iodide with concentrated 
phosphoric acid. 

I- + H 3 P0 4 = HI + H 2 P0 4 

The solubilities of the slightly soluble iodides are generally lower than 
the corresponding chlorides and bromides, but most iodides are fairly 


Reactions (in order of decreasing [I~]) 

PbI 2 = Pb+ 2 + 21- 
Cul = Cu+ + I- 
Agl = Ag+ + I- 
Hg 2 I 2 = Hg 2 + 2 -f 21- 

8.3 x 10- 9 
1.1 x 10- 12 
8.5 x 10- 17 
4.5 x 10' 29 


Iodides are isolated from sea plants and natural saline deposits from 
which the element is obtained by C1 2 displacement. I 2 in alcohol is a 
standard antiseptic and sundry iodides are compounded in medicines. 
Some metals are purified via their iodides, see zirconium in Chapter 20, 
for instance. 

Iodides are determined quantitatively by oxidation to I 2 and titration 
with standard Na 2 S 2 O 3 , using starch indicator, or gravimetrically as Agl. 

O. Thiocyanate (Sulfocyanate) Ion, - SCN~ 

Resonance in this ion shows the possibility of attachment through 
either S or N and both series of salts are known. (See discussion of 
manner of attachment isomers, Chapter 4.) 

f\ .. ^ .. 

- : S O=:N > : S--CN : - 

v_^r " ^_~s 

(thiocyanate) (isothiocyaiiate) 

In water solutions of thiocyanates, however, Raman spectrum shows only 
the triple bonded form. Pure HSCN is prepared by reaction between 
KSCN and KHSO 4 in cold solution, but is of little use. It polymerizes to a 
yellow solid upon heating and in aqueous solution is about as strong an 
acid as HC1. Thiocyanates are affected by heat and reducing or oxidizing 
agents. Products of SCN~ decomposition include C, S, NH 3 , H 2 S, 
NCO-, HCN, SO 2 , SOa 2 , NH+, SOj 2 , CS 2 , COS, CH 3 NH 2 , etc. 

Most thiocyanates are water soluble and the slightly soluble ones 
increase in solubility with excess SCN~ by forming complexes. 


Reactions (in order of decreasing [SCN~]) 

AgSCN = Ag f + SCN- 
CuSCN = Cu+ + SCN- 
Hg 2 (SCN) 2 = Hg+ 2 + 2SCN- 

4 x 10~ 9 
1 x 10~ 12 
4 x 10~ 14 
3 x 10- 20 

With Fe +3 , thiocyanate gives Fe(SCN) 42 (and possibly other ions) 
whose red color is a good qualitative test for either Fe 43 or SCN~. With 
Hg+ 2 , thiocyanate gives a white precipitate of Hg(SCN) 2 which expands 
greatly on heating due to polymer formation (fourth of July "snakes") 
and simultaneous gas release. With Co+ 2 and acetone, SCN~ yields a 


blue cobalt complex (test 17-10), and with Ag + , white AgSCN precipitates. 
This, when heated with dilute HC1 and Fe+ 3 , forms AgCl and the red 
ferric-thiocyanate color. Like S 2 Og 2 and S~~ 2 , SCN~ catalyzes the azide- 
iodide reaction (special experiment 6, part 5) which can also be used as a 
qualitative testing method. 

Thiocyanates are prepared by heating cyanides with sulfur and as a 
by-product in coking. Their uses are limited. 

With regard to quantitative analysis, thiocyanates are gravimetrically 
determined as BaSO 4 (after oxidation to SO^ 2 ), by precipitation and 
weighing of AgSCN, or by colorimetry using the ferric complex. 

P. Ferrocyanide [Hexacyanoferrate(II)] Ion, 

This ion is a mild reducing agent, as ferrous iron is capable of losing an 
electron to yield the ferricyanide, hexacyanoferrate(III), ion containing 
ferric iron: 

[Fe(CN) 6 -*] = [Fe(CN) 6 - 3 ] or + g- E = - 0.36 

These cyano complexes are quite stable (K ins for Fe(CN)g 4 = 10~ 42 ), 
except in the presence of hot alkalis or acids. The latter produce the 
weak acid HCN: 

[Fe(CN)g 4 ] + 6H+ = Fe+ 2 + 6HCN 

H 4 [Fe(CN) 6 ] can be prepared by acidification of an alkali metal ferro- 
cyanide solution, ether extraction, and solvent evaporation. It is a 
crystalline solid which decomposes in solution or moist air to yield some 
Fe+ 3 which reacts with the remaining ferrocyanide to produce a dark blue 
solid called "Prussian blue," KFe[Fe(CN) 6 ] H 2 O. This reaction is 
utilized in blue-print paper, which contains ferric citrate and ammonium 
ferricyanide. Citrate reduces ferricyanide to ferrocyanide, which in turn 
combines with ferric ion when catalyzed by light. 

The first two protons are readily ionizable from H 4 Fe(CN) 6 , whereas 
the last two show weaker acidity: K 3 = 10~ 3 , K 4 = 5 x 10~ 5 . 

Most ferrocyanides are colored and only slightly water soluble, the 
colors of salts of Kg* 2 , Fe+ 2 , Fe+ 3 , Cu+ 2 , Cr+ 3 , Mn+ 2 , Co' 2 , Ni+ 2 , Th lv , 
Ti IV , UO+ 2 , for example, being of value in identification of the cation 
present. K 4 [Fe(CN) 6 ], the most common ferrocyanide, is prepared as a 
by-product from coking. Many double salts like K 2 Zn[Fe(CN) 6 ] are 
made from it and are useful in precipitations for analytical separations. 
With Fe+ 2 and K+ , ferrocyanide yields a white precipitate of FeK 2 [Fe(CN) 6 ], 
which air oxidizes to Prussian blue; with Ag+, ferrocyanide forms white 
Ag4[Fe(CN) 6 ] which is insoluble in ammonia but is converted to red 
Ags[Fe(CN) 6 ] when heated gently with concentrated HNO 3 , and that 



Reactions (in order of decreasing [Fe(CN) 6 4 J) 

K (approximate) 

CaK 2 [Fe(CN) 6 ] = Ca+ 2 + 2K+ + [Fe(CN)^ 4 ] 
Mg(NH 4 ) 2 [Fe(CN) 6 ] = Mg+ 2 + 2NH+ + [Fe(CN) 6 - 4 ] 
Ca(NH 4 ) 2 [Fe(CN) 6 ] = Ca+ 2 4- 2NH+ + [Fe(CNV 4 ] 
BaK 2 [Fe(CN) 6 ] - 5H 2 O = Ba+ 2 + 2K+ + [Fe(CNV 4 ] + 5H 2 O 
MgK 2 [Fe(CN) 6 ] = Mg+ 2 + 2K+ + [Fe(CNV 4 ] 

1.2 x 10~ 2 
9.1 x 10- 3 
8.9 x 10- 8 
6.4 x 10- 3 
6.2 x 10~ 3 

product is ammonium hydroxide soluble. All ferrocyanides are decom- 
posed by heat ; a possible reaction is 

3K 4 [Fe(CN) 6 ] = 12KCN + 1(CN\ + C + Fe 3 C + N 2 

The structure of heavy metal cyanides is still not certain. There is 
evidence that many exist as polymers, chains being built up probably by 
means of the cyano group using an unshared electron pair of N as a 
second valence: 


This is resonance stabilized. Iron uses all its 3/;, 4,y, and 4p orbitals. 

Laboratory Study of Group 3 Anions 

*Test 22-33. C7~ in the Presence of SCN-, Br~ and I". The identi- 
fication of AgCl by color and solub. is difficult in the presence of the insol. 
silver salts of these other ions so they are oxidized before AgCl is pptd. 

(a) Three drops of each test soln. and 1 ml of glacial HAc are mixed in a beaker, 
then 100 mg of PbO 2 is added. The mixt is carefully evapd. in the hood until 
only a moist mass remains. If at this time the vapors still appear violet from 
1 2 or turn starch-iodide paper blue from Br 2 , another half ml of HAc should be 
added and heating and testing continued. When completed, dil. the mixt. with 
1.5 ml of H 2 O, transfer to a tube, centrifuge, and test a little centrate with Ag + 
for C1-. 

(b) If only SCN~ is present, it may be destroyed by simple evapn. of the sample 
in a crucible, baking until charring is over, cooling, then leaching with H 2 O, 
centrifuging, and testing for Cl~, as before. 

Try the solub. of AgCl in dil. NH 4 OH and the result of acidifying that mixt. 
with HNO 3 . Explain with equations; see silver, Chapter 15. 

*Test 22-34. /- and Br~ Oxidations. Iodide is more easily oxidized 
than bromide so a selective oxidizing agent will eliminate I~ which would other- 
wise give reactions similar to Br~. 


(a) Mix 3 drops each of I~ and Br~ solns. with 5 drops of H 2 O. Add 4-6 
drops of CC1 4 , and with shaking add drops of a dil. NaOCl soln. A violet 
color in the organic layer indicates I 2 . Continue with good shaking to add drops 
of NaOCl until the I 2 has been oxidized to colorless IO^. Only a little Br~ has 
been oxidized at this point. Acidify with 6 M H 2 SO 4 and mix cautiously. CI 2 
is released and Br~ is oxidized to Br 2 , and the organic layer shows this with a 
brown bromine color. Write the reaction equation showing iodine, hypo- 
chlorous acid, and water giving iodate, hydrogen ions, and chloride ions. 

(b) Make the I~-Br--CCl 4 mixt. as above. Add a few drops of sodium 
nitrite soln. and acidify with dil. HAc. Explain what happens in words and 
with equations. 

(c) Make the test mixt. as above again. Add 3-^t drops of 3 % H 2 O 2 and a 
drop of 6 M H 2 SO 4 and mix slowly. Does this reagent oxidize only I~ or both 
halide ions ? Give equations. 

+Test 22-35. Br~ in the Presence of I~ and SCN~. (a) Put a few 

crystals or about a ml of sample in a small beaker. Add a ml of H 2 O contg. 
about 50 mg of Fe+ 2 , then add satd. CuSO 4 soln. dropwise until no more cuprous 
iodide ppts.: 2Fe f2 H- 2Cu+ 2 4- 21" = 2CuI + 2Fe+ 3 . Centrifuge and put 
half the centrate in the tube of the Gutzeit apparatus (Fig. 14^4). Mix in a 
few small crystals of K 2 Cr 2 O 7 and 4-5 drops of 6 M H 2 SO 4 and assemble the 
rest of the apparatus as pictured. Put a drop of dil. alcoholic fluorescein on the 
indicator paper (in place of AgNO 3 used in the Gutzeit test) and add a drop of 
50% alcohol to the paper. Heat the tube carefully. Both SCN- and Br~ are 
oxidized but only Br 2 reacts with the reagent and turns it orange pink. See 
example 13, Chapter 12, p. 196. 

14H+ + Cr 2 O^ 2 + 6Br- = 2Cr^ 3 4- 7H 2 O 4- 3r 2 

(b) Preparation of fluorescein (optional). In a 4-in. tube, heat this mixt. 
gently over a low flame for 3-4 min.: 100 mg of phthalic anhydride, 150 mg 
of resorcinol, and 4 drops of coned. H 2 SO 4 . Allow the mass to cool. Add a 
mixt. of 5 drops of 6 M NaOH and a ml of H 2 O, rewarm, and discard the super- 
natant liq. Add a drop of 6 M NaOH and 3 ml of 95 % alcohol and shake. 
The resulting yellow fluorescent soln. is bottled and used as in (a). It should 
be diluted enough to appear light yellow when spotted on the paper. 

Test 22-36. Precipitation and Solubility of AgBr. To 4 drops of Br~ 
soln., add 10 drops of H 2 O, 1 drop of 2 M HNO 3 , and 5 drops of Ag+. Note 
the color of AgBr. Centrifuge, wash, and try the solub. of the residue in dil. 
NH 4 OH and compare with AgCl and Agl (tests 22-33 and 22-37). See also 
example 9-6. 

Test 22-37. Some Heavy Metal Iodides, (a) Repeat test 22-36 using 
I- instead of Br~. 

(b) To a few drops of I~ soln., add a few drops of Hg++ soln. What is the 
red ppt. ? Add more I~, mix, and describe the change. Write the equations. 

(r) Prepare some PbI 2 from PbAc 2 and KI solns. Transfer to a beaker, boil 


with 30-40 ml of H 2 O, and pour the clear supernatant liq. into another beaker. 
After the soln. has cooled and pptn. is complete, view one of the leafy crystals 
under magnification and describe it. Does it appear that some solids could be 
identified by this means alone? 

(d) Try several other cations from the solns. for cation testing and report 
colors of any insol. iodides obtained. 

Test 22-38. Formation of Cuprous Iodide. Add 2 drops of I~ to 
15 drops of H 2 O, then mix in 2 drops of Cu+ 2 soln. and 2 drops of SO^ 2 soln. 
What is the white ppt. ? Br~ and Cl~ do not give this test. 

Test 22-39. Starch Indicator for J 2 . Mix a few drops of I"" soln. with 
a drop of 6 M H 2 SO 4 and a drop of starch solution. Add a drop of some 
oxidizing agent and note the color of the starch-iodine complex. Neutralize the 
soln. with a drop of 6 M NaOH, then add a drop or two of S 2 O^ 2 soln. and note 
the sharp color change as iodine is reduced. These reactions are of prime 
importance in quant, analysis. (See section describing thiosulfate chem.) 
Starch-potassium iodide paper is used to indicate the presence of oxidizing 
agents. Explain how it might be prepd. and what one would observe in a positive 

*Test 22-40. SCN~ in the Presence of Fe(CN)^ and I~. Ferro- 
cyanide and iodide give colors with Fe 13 which obscure the formation of 
Fe(SCN) H 2 , an otherwise sensitive test for thiocyanate. 

Arrange a tube for distillation and distillate collection (Fig. 14-7). Put 3 
drops of each of these three test solns. in the distillation tube, and add a boiling 
chip and 5 drops of coned. HC1. Heat the mixt. and collect the distillate 
(HC1 + H 2 O 4- HSCN) in a test tube contg. 6 drops of 2 M NaOH and standing 
in a beaker of cold water. When several drops of distillate have come over, 
add 3 drops of 6 M HNO 3 to them and a drop of Fe+ 3 soln. A red color 
shows SCN~ in the mixt. The concn. limit is about 0.1 ppm of SCN~. 

Test 22-41. AgSCN and its Solubility. Repeat test 22-36 using SCN~ 
in place of Br~ and compare the results with previous work. 

Test 22-42. Co+ 2 -acetone Test for SCN~. Review test 17-10 and 
decide what modification is needed to make this an analytical method for 
thiocyanate. Write your method in the notebook and give experimental 
evidence that it will work. The concn. limit is about 20 ppm SCN~. 

*Test 22-43. [Fe(CN)^] in the Presence of SCN~. Using the color 
of Prussian blue, one may determine ferrocyanide in the presence of thiocyanate 
in at least two ways: 

(a) Mix 2 drops of each of these test solns. with 10 drops of H 2 O and a drop 
of HC1. Add 3 drops of Fe+ 3 soln., shake, and centrifuge. A red supernatant 
color shows Fe(SCN)* 2 , a blue ppt. is Prussian blue. Ferrocyanide alone may 
be detected at a concn. as low as 3 ppm by this test. 

(b) Make the same mixt. as above but this time add 2 drops of freshly prepd. 


Fe+ 2 soln. A white ppt. gradually turning blue is evidence that an iron ferro- 
cyanide has formed. 

Test 22-44. Heavy Metal Ferrocyanides. Prepare silver ferrocyanide 
using 3 drops of [Fe(CN)JT 4 ], 10 drops of H 2 O, and 3 drops of Ag + soln. Centri- 
fuge and try the solub. in NH 4 OH. Account for the evidences of reaction with 
equations. Repeat using other heavy metal ions as suggested in the description 
of ferrocyanide general chem. As time permits, make several known dilns. of 
potassium ferrocyanide, put 2 drops of each on a different piece of filter paper, 
and add to each a drop of soln. of one or another heavy metal ion. Describe 
the procedure devised and discuss the results. Does this method seem to give a 
way to approx. the concn. of a ferrocyanide soln.? 

Q. Nitrate Ion, NO 3 

The oxidation state of N is (V), the highest that element has, consequently 
nitrates (especially as concentrated HNO 3 ) can act only as oxidizers. 
NO^ may yield, upon being reduced, various products such as NO, NO 2 , 
N 2 O 3 , N 2 O 4 , N a , NHJ, NH 3 , NO 2 , NH 2 OH (hydroxylamine), H 2 N NH 2 
(hydrazine), etc. Some of the redox half reactions involving nitrate are: 

NO + 2H 2 O = NOJ + 4H+ + 3e~ E = - 0.96 

NH+ + 3H 2 = NOg + 10H+ + 8<r E = - 0.87 

N0 2 + H 2 = NOg + 2m + er E = - 0.81 

N 2 + 6H 2 = 2N0 3 + 12H+ + 10e~ E = - 1.24 

The nitrate ion is planar and is stabilized by resonance among three 
equivalent forms; nitrogen oxides are similarly stabilized. Almost all 
nitrates are water soluble. Those of mercury and bismuth hydrolyze to 
precipitate basic salts, as BiONO 3 , but the reactions are reversed with 
dilute HNO 3 . Light metal nitrates give nitrites and oxygen when heated, 
heavy metal nitrates give a metal oxide, nitrogen dioxide and oxygen; 
ammonium nitrate, a special case, decomposes to nitrous oxide and water. 
HNO 3 is prepared by heating Chile saltpeter (natural NaNO 3 ) with 
concentrated sulfuric acid or via the oxidation of NH 3 (Ostwald process). 
The commercial HNO 3 is 68 % nitric acid by weight, but the oxidizing 
power may be increased by solution of nitrogen oxides in it to produce 
white and red fuming nitric acids. These are used as oxidants in liquid 
fueled rockets, being sprayed into a reaction chamber with a combustible 
material like gasoline, turpentine, or aniline. All such mixtures are 
spontaneously inflammable. Mixed H 2 SO 4 and HNO 3 is used in organic 
reactions to introduce the nitro group, NO 2 , into certain organic 
molecules (trinitrotoluene, "T.N.T.") and to make nitrate esters with 


alcohols (glyceryl trinitrate, "nitroglycerine")- Nitro alkanes like nitro- 
methane, CH 3 NO 2 , are used as solvents and in racing fuels. Inorganic 
nitrates such as NH 4 NO 3 are used in explosives and as fertilizers. Four 
tests are described here for the analysis of nitrate. 

(a) A brown ring forms at the liquid junction in a tube containing dilute 
FeSO 4 solution and nitrate on top of a layer of concentrated sulfuric acid, 
due to the presence of nitrosyl iron(II) ions : 

NO 3 + 3Fe+ 2 + 4H+ = 3Fe+ 3 + NO + 2H 2 O 
NO + Fe+ 2 = Fe(NO)+ 2 

Interferences include other reducing, oxidizing, or complexing agents 
such as NOg-, I- Br~ CrO^ 2 , S 2 C 2 , SO^ 2 , SCN~ [Fe(CN)^ 4 ]. These 
can be precipitated, however, with Ag 2 SO 4 . Any remaining nitrite can be 
destroyed with sulfamic acid (see paragraph A, p. 375). 

(b) NO 3 gives a red color with a concentrated H 2 SO 4 solution of 
brucine, one of a class of naturally occurring amines known as alkaloids. 
When the color fades, addition of Sn 11 solution causes a violet color 
to appear. 

(c) With hot H 2 SO 4 and copper, nitrate gives brown NO 2 gas and a 
blue solution of cupric ions: 

Cu + 4H+ + 2NO 3 = 2NO 2 + Cu+ 2 + 2H 2 O 

(d) Nitrate can be reduced in alkaline solution with Devarda's alloy 
(Zn-Al-Cu) or other active metals to yield ammonia: 

8AI + 5OH- + 18H 2 O + 3NO 3 = 3NH 3 + 8A1(OH) 4 

Other nitrogen-containing radicals, Fe(CN)jjr 4 , SCN~, NO 2 , NHJ, must 
be removed first, as they give ammonia also. 

Quantitatively, one measures nitrate content by reduction to NH 3 
as described in (d) above. The NH 3 is distilled and trapped in boric 
acid, then standard acid is used to titrate the NH 4 BO 2 formed. (Winkler 
modification of the Kjeldahl method.) This is possible because boric 
acid is much weaker as an acid than is ammonium hydroxide as a base. 

R. Acetate Ion, H 3 C CO^ (Ac~) 

Acetic acid is stable in water solution as well as in the pure state, 
the latter a water-white liquid that freezes to a solid resembling ice (hence 
the name "glacial" for the pure acid) at 16.6 C. The melting point is 
used as a purity criterion. HAc is a weak monoprotic acid, the H on the 
O being the only ionizable one. Almost all acetates are water soluble. 
(See Chapter 7 for problems dealing with acetate equilibria.) 

When heated with dilute H 2 SO 4 , acetic acid is released from acetates 


and its characteristic vinegar odor detected. If an alcohol is added to 
the mixture, a pleasant smelling ester is distilled off (ethyl acetate here): 

H 3 C-C/ + C 2 H 5 OH = H 3 C-C/ + H 2 O 

X)H H X 0-C 2 H 5 

With hot dilute solutions of Fe +3 , acetate precipitates at about pH 1 as 
basic ferric acetate, a possible formula being Fe(OH) 2 Ac. Distilling an 
acetate with calcium carbonate yields acetone. Acetone in basic 1^ 
solution is converted to triiodomethane, iodoform, whose distinctive odor 
is easily recognized: 

2H 3 C-C0 2 Na c = o Na 2 C0 3 + (# 3 C) 2 C- 
(H 3 C) 2 C--0 + OH- + Ij = HCI 3 + H 3 C C0 2 Na 

Another qualitative test for Ac~ is made by adding La(NO 3 ) 3 , I 3 , and 
NH 4 OH to an acetate solution to precipitate basic lanthanum acetate 
whose surface adsorbs I 2 and gives a blue color. A concentration of 
about 500 ppm Ac~ is needed, which means the test is only about half as 
sensitive as the basic ferric acetate method. Ions that precipitate La +3 
(OH", POj 3 , SOj 2 ) can be removed by pH adjustment to 7 and precipita- 
tion with Ba+ 2 . 

Quantitatively, Ac~ is converted to HAc, distilled to purify, and 
titrated with standard base. 

Laboratory Study of the Group 4 Anions 

Test 22-45. Brown Ring Test for NO 3 . Note in section Q part (a) 
the ions which interfere with this test and remove them accordingly if the soln. 
may contain them. 

Put 2 drops of NOg" soln. in a tube, and add 8-10 drops of H 2 O and 2 drops 
of fresh Fe+ 2 soln. Incline the tube and without homogenizing the mixt., 
let 3-4 drops of coned. H 2 SO 4 run down the tube wall and form a layer on the 
bottom. A brown ring due to FeNO+ 2 at the interface is indicative of nitrate. 
(See the similar test earlier described for nitrite.) 

Test 22-46. Brucine Test for NO 3 . Add a few crystals of brucine 
(POISON!) to 10 drops of coned. H 2 SO 4 and stir to dissolve. On a spot plate 
place a drop of nitrate soln. and add two of brucine reagent. A red color shows 
nitrate (if nitrite has been previously shown absent). When the spot fades to 
yellow, a drop of SnCl^ 2 turns it violet. A blank should be run. A slight 
reddish tint to the reagent is due to nitrate in the sulfuric acid. The exact 
formulas for the nitrated (?) brucine and its tin reduction products are not 
known. This test will detect NOa" in concns. as low as about 2 ppm. 

*Test 22-47. Reduction of NO^ to NH 3 and to HNO 2 . Note the inter- 
ferences to this test as explained in paragraph Q, p. 404. 


(a) To 50 mg of Devarda's alloy add 6 drops of nitrate soln., 4 drops of 
6 M NaOH, warm, then carefully note the odor of NH 3 . Write the equation 
using Zn as the reacting metal. How could Nessler's reagent (test 19-9) be 
utilized? Invent a method and report on it. 

(b) If NOg~ is present, remove it via test 22-1 (b). Make soln. neut. with 
NH 4 OH, then slightly acid with HAc. Make a mixt. of a drop of a-naph- 
thylamine, a drop of sulfanilic acid, and a little Zn dust. Add the nitrate soln. 
to this and agitate it now and then during the next few min. A very definite color 
indicates NO^ was in the sample, since it has been reduced to HNO 2 which 
gives the Griess reaction (test 22-5). 

*Test 22-48. Acetate in the Presence of Interfering Ions, (a) When 
one attempts to identify Ac~ in the presence of group 1 anions, the odor tests on 
HAc and its esters fail because other volatiles like NO 2 and SO 2 mask them. 
One may ppt. all the ions except Ac~ with excess AgNO 3 , then proceed with the 
centrate to test for acetate. In this method some nitrite escapes although the 
excess pptng. agent decreases the solub. of AgNO 2 and at the same time prevents 
the formation of the soluble silver thiosulfate complex, Ag(S 2 O 3 )jjT 3 . Treatment 
of the centrate with sulfamic acid (see test 22-1 and paragraph A p. 382) destroys 
NO 2 traces. The student should try this procedure on a known mixt. contg., 
for instance, SO.^ 2 , NO^, and Ac~. 

(b) Another procedure is to oxidize all the group 1 ions except Ac~ and CO^ 2 
in a warm neutral or slightly acidic soln. with KMnO 4 . One may then test 
the centrate for Ac~ by heating with H 2 SO 4 to get the HAc odor or by one 
of the tests below. The student should also try this procedure on a known 

Test 22-49. Basic Ferric Acetate. Put 4 drops of neutral Ac~ test soln. 
in a test tube, add 10 drops of H 2 O, 3 drops of FeCl 3 , and boil briefly. A red- 
brown color or ppt. is ferric hydroxy acetate. Interferences include S~ 2 , SOs" 2 , 
SCN~, Fe(CN)jT 4 , and I~, but they can be pptd. with Ag+. The concn. limit 
is about 250 ppm Ac~. 

Test 22-50. Conversion of Acetate to Acetone and to Esters, (a) Heat 
a mixt. of about 100 mg of solid CaCO 3 with 20 drops of Ac~ soln. in a test 
tube and slowly evap. it to 2-4 drops. Connect a delivery tube and arrange 
to collect the distillate in another tube standing in a 50 ml beaker of cold H 2 O 
(Fig. 14-7). Evaporate the mixt. and heat it strongly. To the distillate of 
acetone and water add a drop of 6 M NaOH and a drop of KI-I 2 soln. and shake. 
The appearance of a fainjt yellowish ppt. of iodoform accompanied by its easily 
recognized medicinal odor is a positive test for acetate. The sensitivity is about 
the same as for the preceding test. Group 1 anions should be removed prior 
to making the test. 

(b) Acetic acid forms pleasant-scented esters when caused to react with 
alcohols, in the presence of a catalytic amount of mineral acid. To 10 drops 
of Ac~ soln., add 5 drops of an alcohol, a drop of 6 M H 2 SO 4 , and warm. Note 
the fruit-like odor of the ester. Write the equation. 


Test 22-51. Lanthanum-Iodine Test for Acetate. Put 2 drops of 
Ac- test soln. on a spot plate, add a drop of 5% lanthanum nitrate, a drop of 
0.01 N IJT (I 2 in KI) and a drop of 2 M NH 4 OH. Depending upon the [Ac~] a 
blue color or ring develops. The concn. limit is about 500 ppm. 


(1) If the sample is known to be only a mixture of soluble Na + and K+ 
salts of the 18 anions just listed, one should test small portions of it with 
the three group reagents (hot dilute H 2 SO 4 , Ca+ 2 -Ba+ 2 , and Ag 4 ) to 
determine if tests on individual ions within those groups are necessary. 
Group 4 ions, NO^ and Ac~, must be tested for individually. (It is 
assumed that explosively powerful oxidizing ions as ClO^ , CIO\ , and 
5 2 Og 2 are absent.) After deciding in which groups to make tests, one 
systematically runs individual tests, as described, using a fresh small 
portion of unknown each time unless a sequence is specified. Cognizance 
of interferences must be taken and such substances removed. Sometimes 
the first test for an ion will be conclusively positive or negative, sometimes 
several are needed. The tests marked * listed after the description of the 
ion are recommended as the ones most likely free of difficulties. Where the 
words "test solution" or "solution of that ion" appear, one substitutes the 
words "unknown solution," and. proceeds as if repeating the preliminary 

(2) If the sample is soluble but cations other than Na+ and K* are suspected, 
they may interfere with anion tests. Such cations form insoluble car- 
bonates and can be not only detected but eliminated in that way. Para- 
graph Y in this chapter, p. 411, gives the procedure. 

S. Testing Anion Group 1 

If the unknown is a soluble solid, dissolve about 200 mg in 5 ml of 
H 2 O. (The solution should not be too concentrated since the tests are 
designed for small quantities.) Test the solution's pH by putting a drop 
on a spot plate and adding a drop of phenolphthalein ; since group 1 
anions hydrolyze, it is characteristic of their alkali metal salt solutions to 
have a high/?H (PO^ 3 , BO 2 , CgOj 2 , and Ac~ give the same reaction also). 
To several drops of solution, add 1-2 drops of 6 M H 2 SO 4 and note the 
appearance of gas evolution. Warm and again observe the result, 
cautiously getting the odor above the tube. If the solution fizzes immedi- 
ately and the gas is odorless, a carbonate or bicarbonate is probably present. 
The odor of SO 2 indicates a sulfite or bisulfite is present and a simultaneous 
fine yellow-white precipitate may be sulfur and indicates the possibility of 
a thiosulfate. The odor of H 2 S is unmistakable and shows sulfides or 


bisulfides and a brown gas is most likely NO 2 , created when a nitrite 
decomposed. (Some HAc odor indicating acetate may be noted here if 
other odors are faint or missing.) If any of these evidences of group 1 
anions is apparent, individual testing of appropriate ions according to the 
preliminary exercises is undertaken. This solution can be used to 
examine group 3 below (decomposition of NO^ gives NOj in solution), 
if heating is continued to evolve all easily produced gases. If no gas 
evolution is noted, try the test again using concentrated HC1 on some 
solid sample. The absence of gas release indicates the absence of group 1 
anions, though nitrite should be checked by an independent test, and 
oxidizing agents may release C1 2 from the HC1. 

T. Testing Anion Group 2 

To another small volume of unknown solution add 2 drops of 6 M 
HC1 and warm the mixture to decompose and expel the decomposition 
products of group 1 ions. If no group 1 was present, HC1 addition is 
omitted. Neutralize any HC1 added with a drop or two of NH 4 OH, 
add a drop of methyl orange and enough 2 M HAc to change the indicator 
red, and then add 3 drops more of HAc. Add a few drops of CaCl 2 
solution; a white precipitate indicates the probable presence of F~ 
and/or C 2 O 4 2 , and no precipitate indicates their absence. The solid cap 
be dissolved in HC1 and tested for the two anions, remembering that 
C 2 O 4 2 must be destroyed before F~ is analyzed colorimetrically. 

Treat the centrate with just enough 3 M NH 4 OH to turn the indicator 
yellow then heat the tube in a water bath and add several more drops of 
CaCl 2 and several of BaCl 2 . A yellow precipitate indicates CrO 4 2 , a 
fine white precipitate possibly shows SO^ 2 , and bulkier white precipitates 
may be the barium salts of BO 2 or PO 4 3 . Add 4 drops of dilute HC1 and 
reheat, with stirring. If a precipitate remains it is BaSO 4 , if it only slowly 
dissolves, it may be Ba 3 (PO 4 ) 2 , If no precipitate formed with the barium- 
calcium mixture, then appreciable amounts of PO^, SO 4 2 , CrO 4 2 , F~, 
C 2 O 4 2 , and BO 2 are absent. If only small quantities are suspected present 
then those tests which are most sensitive should be tried. Of these 
residues, borate is most soluble and must be tested for even if no group 2 
precipitate comes down. 

U. Testing Anion Group 3 

The solution from the H 2 SO 4 testing of group 1 is centrifuged if any 
residue is present, and the centrate examined further. To a portion, add 
2 drops of 6 M NH 4 OH, 5 drops of 6 M HNO 3 , and enough Ag+ to com- 
plete precipitation of group 3 ions. Group 2 anions will not yield 
insoluble silver salts in the presence of HNO 3 but group 3 ions will. A 


yellow precipitate is probably Agl, a white or faintly yellow precipitate 
may be any or all of the silver salts of Cl~, Br", SCN~, and [Fe(CN)^ 4 ]. 
Try the solubility of the precipitate in 0.3 M NH 4 OH. If it readily 
dissolves and SCN~ is shown to be absent, Cl~ was probably the only 
anion present (though test 22-33 should be used to check). Addition of 
dilute HNO 3 to that solution reprecipitates white AgCl. The other ions 
are tested in the remaining portion of solution from group 1, using 
tests 22-34, 22-35, 22-40, and 22-43 on separate parts. 

V. Testing Anion Group 4 

No general reagent is available for this. Groups 1, 2, and 3 are first 
eliminated, then group 4 tests are made. To a small portion of unknown 
solution add Ag 2 (SO 4 ) solution to complete precipitation of all ions 
giving insoluble silver salts. Centrifuge. Treat the centrate with the 
CaCl 2 -BaCl 2 mixture to complete precipitation of other ions (except 
Oh and group 4). Centrifuge and test the centrate for Ac~ and NOg 
as described in tests 22-47 and 22-49. These tests are not obscured by 
Cl~ introduced with the calcium-barium solution. 

W. Testing for oxidizing Agents 

If strong oxidizing anions are present in the unknown then strong reducing 
anions are absent. The extent to which these react with one another is 
chiefly dependent upon the distance of separation of their oxidation 
potentials and the solution /?H; in general a considerable difference in E Q 
values and a low pH give optimum reaction conditions. Thus at high 
/?H, S(>3 2 and S~ 2 do not react with each other, but if the mixture is 
acidified, sulfur and water result. One checks for oxidizing and reducing 
ions, consequently, not only to find what is in the sample but also to get a 
clue as to what is not. A negative test can be almost as helpful as a 
positive one. 

To 4 drops of unknown solution, add 2 drops of 12 M HC1 and 6 drops 
of saturated MnCl 2 in HC1, and warm the tube in a water bath. A brown 
precipitate of MnO 2 indicates that NO 2 , CrO^ 2 , and/or NOg are in the 
sample. After the residue settles, the supernatant solution may be green 
due to Cr+ 3 , from a chromate, or yellow, due to dissolved nitrogen oxides 
from a nitrite or nitrate. Nitrate gives the poorest response in this test 
and should be checked again as described in paragraph V. 

X. Testing for Reducing Agents 

If oxidizing agents were not present, reducing ions may be. 
To 6 drops of sample solution, add 4 drops of H 2 O and 2 drops of 
6 M H 2 SO 4 . Mix and cool. Now add 2 drops of 0.01 M KMnO 4 , 


A bleaching of the color shows that at least one of these is present: SOg 2 , 
I~, Br~, S~ 2 , NO^, and SCN~. No bleaching at room temperature or 
below means that these ions are probably absent. Continue adding 
oxidizing agent with mixing until a definite red color remains, Now 
heat the tube in a water bath. If more bleaching takes place, CgO^ 2 
is probably the cause, or appreciable quantities of Cl^ are present. Test 
further for these ions if the reaction was positive. 

A further useful test for reducing agents is described by Lundin (refer- 
ences). Drops of solution are put on starch-iodate paper and the paper 
turns blue if reducing ions are present which can act on iodate to liberate 
iodine. Reported sensitivities in mg/ml of the ions are 

S- 2 0.1, SOg 2 0.1, S 2 (V 10, [Fe(CN)^ 4 ] 10, r 0.1, NO 2 ~ 10, SCN~ 0.05 

Y. Insoluble Samples For Anion Analysis 

If the unknown is not already in solution and gives evidence of slow or 
low solubility in H 2 O and dilute acids, a portion should be powdered in an 
etched spot plate depression, using a blunt stirring rod as a pestle. A 
little of this material is tested for carbonate using test 22-15, then the 
rest is transposed by boiling with sodium carbonate solution or by fusion 
with the solid in a platinum crucible. For example: 

Ba 3 (P0 4 ) 2 + 3Na 2 C0 3 4 2Na 3 PO 4 + 3BaCO 3 

In this way the anions are put into solution, since the sodium salts formed 
are water soluble. 

With stirring in a 20-ml beaker, boil a mixture of 300 mg of sample 
with 1200 mg of cp anhydrous Na 2 CO 3 and 8 ml of H 2 O, or use propor- 
tionate amounts on available sample. Continue to keep the mixture hot 
for at least 10 minutes, and replace water as it evaporates. Rinse the 
mixture into two tubes, centrifuge, and save both centrates and residues. 
Wash the residues and combine washings with centrates. Use small 
portions of the centrate and make individual ion examinations as already 

The residue may contain carbonates and hydroxides of the first four 
cation groups, plus Mg+ 2 , as well as some untransposed F~, S~ 2 , Cl~, Br~, 
I", and PO^ 3 . If these ions are found in the centrate, they need not be 
analyzed again here, since enough was solubilized for testing purposes 
before. If the residue contains carbonates only, it should be soluble in 
warm 3 M HAc and if this is found, then only the centrate above requires 
anion analysis. If some of the residue remains after HAc addition, 
portions are used as directed below. 

1. Fluoride. To a portion of residue contained in a 4-inch test tube, 


add, in order, a few granules of powdered glass, 2 ml of 6 M H 2 SO 4 , 
and finally 25-50 mg of solid Ag 2 SO 4 , and arrange to distill (CAUTION!) 
and collect the distillate. The silver salt precipitates some of the ions and 
prevents their distillation as volatile acids. The glass reacts with HF 
giving H 2 SiF 6 , which distills. Collect about 1 ml of distillate in a tube 
cooled in a beaker of water and test it colorimetrically for fluoride. Com- 
pare with a blank and a known. See test 22-27. 

2. Sulfide. To a separate portion of residue, add a few mg of Zn 
and a ml of 6 M HC1. Test for H 2 S release by PbAc 2 paper. 

3. Halides of silver. To a separate portion of residue, add 2 ml of 
3 M HNO 3 and boil. Cool and centrifuge. Save the centrate for part 
4, below. To the residue add 1 ml of 0.3 M H 2 SO 4 , 6-10 drops of 
CH 3 CSNH 2 , and heat in the water bath. Silver halides are converted to 
black Ag 2 S and soluble halide ions. Centrifuge. The centrate contains 
the halides, sulfate, and acetic acid (from thioacetamide). Cl~, Br~, and 
I~ are then tested by procedures given in this chapter, such as tests 22-33, 
22-34, and 22-35. 

4. Phosphate. To the centrate from 3, above, test for phosphate with 
molybdate and nitric acid (test 22-16). 

Z. Further Aid in Anion Analysis 

If the analysis of the cations in the sample has been made prior to anion 
analysis, considerable aid is given the analyst in the table of solubilities 
(back cover), for he can eliminate some cations and suspect the presence 
of others by solubility behavior of the sample. This gives the student the 
opportunity to use his ingenuity, acquired knowledge of chemical facts, 
and application of principles. 

The analysis of the general unknown for cations is given in Chapter 21, 
including treatment of slightly soluble samples. Note there also the 
information to be gained by dry ignition of the sample, as some of the 
observations apply directly to anions. 


1 . Use the table of oxidation potentials to answer the following. 

(a) A solution of sodium nitrite and potassium chromate is acidified. What 

(b) Can H 2 O 2 oxidize Br~ to Br 2 in acid solution ? 

(c) Will sulfite bleach permanganate at low /?H ? 

(d) In acid solution can MnOj" oxidize I~ to IO^", or only to I 2 ? 

(e) Will Cr 2 Of 2 selectively oxidize 1~ and not Br~? 


(/) PbO 2 is heated with hydrochloric acid. Is chlorine gas released? 
(g) In basic solution, can zinc metal reduce Fe(OH) 3 ? 
(h) May Fe+ 3 be used to selectively oxidize I" and not Br~? 
(/) H 2 2 + H 2 Cr 2 7 = 

(/) Why does addition of chlorine water (C1 2 in H 2 O) to a mixture of KBr in 
H 2 O and CC1 4 give a brown CC1 4 layer? 

2. (a) Why is I 2 essentially insoluble in H 2 O but quite soluble in aqueous KI ? 

(b) Why is I 2 in KI solution brown, but in CC1 4 , purple? 

(c) What color would you expect I 2 to be in CH 3 OH solution? In CS 2 ? 
In chloroform, CHC1 3 ? Why? What color is Br 2 in each solution? 

3. What simple test(s) will distinguish between the following pairs if each is in 
a different container? 

(a) Agl and AgCi (b) CaCrO 4 and BaCrO 4 (c) AgCl and AgSCN (d) NaNO 3 
and NaN0 2 (e) H 4 [Fe(CN) 6 ] and HSCN (/) Ag 4 [Fe(CN) 6 ] and AgBr (g) 
Na 2 S 2 O 3 and Na 2 SO 3 (h) K 2 SO 3 and K 2 SO 4 (/) (NH 4 ) 3 ?O 4 and NH 4 NO 3 (y) 
CaF 2 and CaC 2 O 4 . 

4. A general unknown is a mixture of white salts and is water soluble. It is 
found to contain Pb +2 , Ba +2 , and Na+. Which negative ions are probably not 
present in appreciable quantities? 

5. A general unknown is water soluble and found to contain Cl~, SO 4 2 , and 
S~ 2 . What cations should one test for? 

6. A white solid is a single simple salt; it is not soluble in water but reacts 
with H 2 SO 4 to give an odorless gas and another white residue. What might the 
sample have been ? Explain. 

7. What anions are possibly present in moderate quantity in a water soluble 
unknown which is analyzed first for cations and found bearing Ag H , Ca +2 , Ba +2 , 
NH+ , Li+, and K+ 

8. A certain water soluble sample is partially analyzed and Br~, SO 4 2 , NO^, 
and PO^ 3 are found. What cations and anions are probably not present? 
Give a reason for each choice. 

9. (a) A solution of FeSO 4 that has been on the shelf for some months gives a 
dark red color when KSCN is added. Give a theory to account for this. 

(b) A solid mixture containing NaNO 2 and Nal slowly turns dark. Explain. 

(c) An old sample of NaNO 2 gives the brucine test. Explain. 

10. J. P. Slipshod offers helpful hints on handling a solution that may contain 
any combination of these ions: Cl~, Br~, SO 4 ~ 2 , SCN~, NO 2 , and SOa 2 : 

"Divide the solution among six tubes and proceed with six spot tests to 
blanket the entire problem: 

Tube 1 : add H 2 SO 4 and heat ; a gas release proves sulfite. 

Tube 2: add Ag+; a white precipitate proves chloride. 

Tube 3: add Fe+ 2 ; a red color is thiocyanate. 

Tube 4: add chlorine water, H 2 SO 4 and CC1 4 ; a brown upper layer proves 

Tube 5: add a KMnO 4 crystal; decolorization shows nitrite. 

Tube 6: add Ba+ 2 solution; a gelatinous, white precipitate proves sulfate." 
Which points in the Slipshod methods are specious? 


11. Balance and decide in which direction these go: 

(a) H 2 Cr 2 O 7 + Br~ = 

(b) H+ + Cl- + lOg = 

(c) I 2 + Fe(CN)^ 4 = 

(d) ClO^ + H 2 S0 3 = Cl- + H+ + HSOJ 

(e) Cr+ 3 + S + H 2 S~ 2 + CrO, 2 -f H+ 

12. One runs the transposition reaction according to paragraph Y, p. 41 1, 
with 10 ml of 1 M Na 2 CO 3 on solid CaC 2 O 4 . 

(a) Find the ratio [CgCWtCOg 2 ] at equilibrium with Ca+ 2 . 

(b) What weight of CaC 2 O 4 is transposed if COg 2 is added as fast as it is used 
up, and [COa 2 ] is maintained at 1 M ? 

13. Air contaminated with NO 2 (nitrogen dioxide) may be stripped of that 
component by bubbling it through aqueous base or even plain water: 

2N0 2 + H 2 = HN0 2 -f H+ + NO^ K^ ^ 10 5 

You are called in by an air pollution commission to devise a sensitive method 
for the detection of NO 2 in the air. Explain your complete solution to this 
problem including sketch of an air sampling device. How might the method 
be made quantitative? 

14. (a) Boron at a concentration as low as 2-5 ppm in irrigation water is 
known to kill citrus trees. You are asked to recommend a method or methods 
capable of indicating borates under these conditions. Could a person with no 
chemical background be taught to run the test in the field? 

(b) (Library) Chromates are sometimes added to industrial water as corrosion 
preventatives for iron pipes. Explain how this might work. Look up the 
diphenylcarbazide method and summarize its use as a Cr yi reagent including 
equation, sensitivity, and evidence of positive reaction. 

(c) Sulfites are sometimes added to industrial water to prevent iron corrosion 
by acting as an "oxygen scavenger." Explain. You are asked to devise a 
test capable of detecting small amounts (~ 10 ppm) of residual sulfite in the 
water. The water contains only the other usual ions. What may happen if 
considerable sodium sulfite is added to a recirculating water system over a long 
period of time? 

(d) A wine sample is dark red and tastes particularly sour. How could you 
test it for acetic acid ? 

(e) Someone proposes a method for cutting up a certain variety of sea weed, 
boiling it in water, filtering, evaporating the filtrate, and getting iodide salts from 
it. You are asked for an iodide test. Considerable NaCl is known present. 
What is your test? 

15. (0) Copper pot cleaners frequently contain oxalic acid, which forms 
soluble copper complexes, and an abrasive, such as, diatomaceous earth, which 
is essentially SiO 2 . How would you proceed to prove that a sample corresponded 
to this formulation ? 

(b) A solid, white mixture put out for the purpose of silver plating brazed 
joints in electric circuits, contains AgCl, NaCl, NaHCO 3 , and bentonite (clay). 
How would one analyze the product? 


(c) A green-colored solid mixture which is sprinkled on fireplace fires to 
produce colored flames is made from CuSO 4 -5H 2 O and KC1. Why isn't 
CuSO 4 -5H 2 O used alone, or why not use CuCl 2 ? How could the mixture be 
analyzed? Could one use NaCl in place of the more expensive KC1? 

(d) A solid, grey mixture, meant to be sealed in plastic bags with shaved ice, 
is on the market for incorporation in short air shipments of perishable items. 
The product contains asbestos fiber, fuller's earth (an absorbant clay), and 
potassium chloride. What is the function of each ingredient? How would 
one analyze the sample? How would a microscope aid the investigation? 

(e) Restaurants may cut up potatoes for frying several days in advance, but if 
not properly preserved, they turn dark or become slimy even in the refrigerator. 
An aqueous solution sometimes used to keep potatoes white contains NaHSO 3 , 
NaHCO 3 , and citric acid. How could one analyze a sample? 

16. (a) A solution is suspected of containing S 2 O^ 2 , PO^ 3 , Br~, and SCN". 
Give two methods for analyzing the sample. 

(b) An unknown solution gives no gas when heated with dilute H 2 SO 4 . It 
gives a white precipitate with the Ca+ 2 -Ba+ 2 reagent, and the precipitate is 
insoluble in 6 M HC1. The sample also gives a white precipitate with Ag+ plus 
HNO 3 . This is soluble in NH 4 OH and if that solution is acidified with HNO 3 
and then treated with chlorine water and CC1 4 , no reaction is apparent. What 
anions might be present ? 

(r) To a few mg of Ag 2 CrO 4 , one adds a 2 M NH 4 C1 solution. Explain 
what happens and why. If 2 M NH 4 NO 3 were added instead, what would tne 
result be? (Hint: consider the chromate-dichromate equilibrium and the 
hydrolysis of the ammonium ion.) 

17. (a) With respect to the base F~, which acts as the stronger acid in aqueous 
solution: HForH 2 O? Explain. 

(b) Calculate the concentration of each of these in 0.5 A/HF: H+, F~, HF, 

18. (a) Organic chemists have found that when esters are formed according 
to the reaction described in the section on acetate (p. 406) the oxygen in the 
water formed came from the acid and not the alcohol. How could one prove 

(b) In making chromate precipitates like BaCrO 4 , laboratory directions usually 
specify approximately neutral solutions to which are added sodium acetate. 
Why does this procedure insure a good yield of the chromate? 

19. Tell what complications one encounters in the following and how they 
are circumvented : 

(a) Analyzing Ac~ in the presence 

(b) Detecting SO^ 2 in the presence of S 2 Og 2 and SO^ 2 . 

(c) Finding Br~ in a mixture of I" and Cl~. 

20. (Library) Common anions sometimes tested in a beginning course in 
addition to those mentioned in this chapter include CN~, ClO^, C1O~, AsO 4 3 , 
and C 4 H 4 Og 2 (tartrate). Find a good test for each and describe it, including 
interferences, elimination of interferences, and sensitivity, if possible. 


21. (a) Why is the azo dye test for nitrite better than the test involving 
bleaching of permanganate? 

(b) Why is the acetone-iodofomn reaction better for acetate detection than 
the ester-odor test? 

(c) Why is the zirconyl-alizarin test superior to the etching test for fluoride 

22. In one test for carbonate and bicarbonate, acid is added to the sample 
and the gas evolved is bubbled through a dilute Na 2 CO 3 -phenolphthalein 
solution. The fading of the indicator color is taken as evidence of a positive 
test. Explain, including equations. 

23. C 2 O^ 2 and F~ both give a yellow color with the alizarin-zirconyl fluoride 
reagent. Suggest how to rid a solution of oxalate without distillation so one 
can test for fluoride. Include equations for destruction of oxalate, remembering 
that the solution prepared for F~ determination must not contain any colored 
ions which might mask the F~ test. 

24. Which of the following oxidizing agents in acid solution will oxidize I~ 
but not Br~? Which will oxidize I~ and Br~ but not Cl~? Which is/are 
capable of oxidizing all these halide ions? (a) F 2 (b) HNO 2 (c) O 2 (in going to 
H 2 O) (d) Cr 2 O7 2 (e) Fe+ 3 (/) MnOj (in going to Mn+ 2 ). 

25. The solubility of lead thiocyanate is about 0.45g/liter. Place it in 
Table 22-14. 

26. (d) Various nitrate reduction products are listed in paragraph Q. What 
is the oxidation state of N in each? 

(b) Give balanced equations showing what happens when the following are 
heated: NH 4 NO 3 , NH 4 NO 2 , NaNO 3 , Pb(NO 3 ) 2 , HNO 3 . 

(c) One has a mixture of NH/ and NOg". How could it be analyzed for each 
ion quantitatively by the Kjeldahl method? Give equations and explain. 

27. Note the interferences in test 22-49. Explain with equations why each 
hinders the test. 

28. A certain ink for marking glassware contains ammonium bifluoride and 
barium sulfate. Explain. How could one analyze it? 


1. J. S. Pierce and E. Hazard, J. Chem. Educ., 21, 126 (1944). (Ag group) 

2. R. C. Brasted, /. Chem. Educ., 28, 592 (1951). (NO 2 and sulfamate) 

3. F. R. Lowdermilk, R. G. Danehower, and H. C. Miller, /. Chem. Educ., 28, 246 
(1951). (F a and fluorides) 

4. J. A. Lundin, /. Chem. Educ., 28, 122 (1951). (Starch-iodate paper) 

5. D. Reilly, /. Chem. Educ., 30, 234 (1953). (N compounds) 

6. C. A. Noll, Anal. Chem., 17, 426 (1945). (NOg-brucine) 

7. W. R. Crandall, Anal. Chem., 22, 1449 (1950). (F~ quick test) 

8. L. Silverman and K. Trego, Anal. Chem., 25, 1264 (1953). (B colorimetrically) 

9. J. Bergerman and J. S. Elliot, Anal. Chem., 27, 1014 (1955). (H 2 C 2 O 4 colori- 

10. M. M. Markowitz, Anal. Chem.. 33, 36 (1956). (Condensed phosphates) 



The following experiments are roughly divided into two types: the 
first four illustrate principles such as equilibrium constants, /?H, and 
buffers; the last nine are practical applications of qualitative analysis 
and an introduction to the use of siich analysts' tools as the spectroscope, 
ultraviolet radiation, electrography, chromatography, etc. These experi- 
ments may be used as lecture demonstrations or given to students as 
extra-credit material, or simply assigned to the class as regular laboratory 
exercises. It is felt that this section broadens the student's concept of the 
area of chemical analysis and serves as a summation to the course with its 
human interest value, practicality, and spirit of discovery. Related 
topics for general interest reading are listed in a short compilation of 
references following the experiments. 


Investigation of a Typical Weak Electrolyte 

1. The degree of ionization or dissociation of a weak electrolyte in 
water solution may be measured by three methods discussed in previous 
chapters: (a) from freezing and/or boiling point data (6) from voltages 
of properly constituted cells, and (c) from measurement of electric con- 
ductivity. The last method as described below is simple, rapid, and 
productive of good results. 



A conductivity cell and laboratory-constructed Wheatstone bridge cir- 
cuit are satisfactory for the experiment, or one may use a commercially 
available bridge and dipping cell.* 

From the electrical resistance of a solution one may calculate its 
conductance and, from the known concentration of the solution, the 
solute's equivalent conductance. From handbook data and application 
of Kphlrausch's law one may find the equivalent conductance of the 
solute at infinite dilution, and, by comparison of this with the value 
obtained experimentally, the solute's degree and percent of dissociation 
can be calculated. The calculation is illustrated in Chapter 5. 

2. Assemble the particular apparatus as instructed. 

3. Various student groups may be assigned different concentrations of 
some weak electrolyte on which to gather data and make calculations. A 
table of class data can be then assembled on the board for copy in the 
laboratory notebook with comments on the experiment and answers to 
the questions. A suitable system for study is acetic acid-water in the 
concentration range 0.4-0.0001 M. 

If solutions are not already prepared at the various concentrations, 
make 250 ml of the particular concentration assigned from a stock bottle 
of HAc and distilled water, using a cylinder or pipets and a volumetric 
flask as directed. If small conductivity cells are employed, smaller 
volumes are used. Adjust the temperature of the solution to some 
predetermined value like 25 C so the class data will be consistent, rinse 
the cell twice with small volumes of the solution, then measure the resis- 
tance of the solution and calculate the other values. At 25 C, A for 
HAc = 372 (see Table 5-2). The cell should be rinsed with distilled 
water and the apparatus made available for the next group. Always 
treat the cell and bridge carefully. 

4. Copy the class data from the board. From your individual measure- 
ments, calculate and contribute the following to the table: (a) Molarity 
of solution, (b) Specific resistance, (c) Specific conductance, (d) 
Volume in ml needed to contain 1 g equivalent weight of solute, (e) 
Equivalent conductance (from your data). (/) Equivalent conductance 
at infinite dilution (from reference data), (g) Degree of ionization. 
(h) Per cent of ionization. (/) Theoretical per cent of ionization as 
calculated from the book value of the ionization constant. (/) K lon 
(experimental), (k) Plot a graph of per cent ionization versus concentra- 
tion from the entire data collection, and explain. 

* A good instrument for this purpose is the type RC conductivity bridge manufac- 
tured by Industrial Instruments, 1 7 Pollock Ave., Jersey City 5, New Jersey. The 
dipping cell should have a cell constant of about unity, and should be given by the 
instructor to save lab time. 





25 C 

x 10- 4 

wt. x 10 3 






























































5. Questions: (/) Why does dilution change the per cent ionization 
butnotA;. on ? 

(m) Which contains more H + and Ac~ ions: 1 liter of 0.01 M HAc 
or 1 liter of 0.001 M HAc? Show calculations. 

(ri) What factors affect the cell constant? Why doesn't one always 
assume its value is unity ? 

(6) Could one readily check the last table value for per cent ionization 
by a freezing point method? Explain. 

(p) Draw a diagram of the cell used. Label parts. 

(q) Draw and briefly explain a Wheatstone bridge circuit. For what 
is it specifically used here? (See any physics text.) 

(r) What results would you expect by conducting the same experiment 
on NH 4 OH solutions to determine the degree of dissociation at different 

(s) Conductivity measurements are used industrially to check such 
things as the extent of carbonation in soft drinks, purity of distilled water, 
operation of ion exchangers and the mineral content of boiler feed waters. 
Explain how this simple, rapid method of quality testing applies. 

Investigation of Typical Strong Electrolytes (Optional) 

6. Make several dilutions of HC1 in the range 0.2-10" 3 M, from a 


stock solution of dilute acid. Repeat the experiment and calculate parts 
(a) through (k) in paragraph 4. Explain the differences noted. 

7. Make several dilutions of some salt like KC1 in the range 0.2-10~ 4 M 
and measure the conductivities. Prepare a graph of specific conductance 
versus concentration and extrapolate to zero concentration (infinite dilution) 
to find the limiting equivalent conductance. Check this answer by 
applying Kohlrausch's law to appropriate handbook data. 

8. Measure the conductivity of solutions of each of the following at the 
same temperature and concentration, say 25 C and 0.02 M : KC1, K 2 SO 4 , 
MgCl 2 , MgSO 4 , Pb(NO 3 ) 2 , BaAc 2 , and PbAc 2 . Compare these figures. 
Explain why the last salt gives a figure greatly different from the others. 
How does this help to explain why, in cation group 2 analysis, PbSO 4 
is soluble in NH 4 Ac? 

See references at the end of Chapter 5. 


If a pH meter is to be used, its construction and operation will be 
briefly discussed by the instructor. If indicators are to be used, a list of 
them similar to appendix A 17 will be given. A series of buffer solutions 
(see a handbook) will be furnished with the indicators. Since various 
meters may be used, no effort is made here to give specific operating 

A. Finding K A for a Weak Acid 

1. Various student groups may be assigned the task of preparing 
solutions of a weak acid like HAc as in special experiment 1. More or 
less solution will be needed depending upon the method used but 50 or 
100 ml is satisfactory. 

(a) If a meter is employed, adjust the temperature of the solution to 
25 C and, after balancing the meter with a suitable buffer or standard,* 
measure the pH of the acid solution, where this dissociation has taken 

HA = H+ + A- 

If one knows the concentration of H+ (by conversion of the/>H reading), 
he may assume the A~ concentration is the same, and the HA concentra- 
tion is the original solution molarity minus the quantity dissociated as 

* Two solutions for/?H reference at 25 C are 0.030 M potassium hydrogen tartrate, 
pH 3.57, and 0.05Q M potassium hydrogen phthalate, pH 4.00. pU does not vary 
much with small temperature changes. 



represented by either [H+] or [A"]. This is enough data to calculate K 4, 
or with K A given to calculate the per cent dissociation as in special 
experiment 1. Class data may again be assembled from group contribu- 
tions. Note that this instrumental measurement has again led to a means 
for determining such fundamentals as ion constants and dissociation 
percentages. Table 23-2 is data gathered by this method. 

(b) If indicators are used, trial and error experimentation is used to 
get the pH roughly, and a more accurate approximation is obtained by 
comparing the hue of the particular indicator with the sample and with 
buffers in the same pH range. A "universal indicator" can also be 
employed as another check. (See the reference below.) 



P H 
25 C 

M x 10~ 3 

Per Cent 

Per Cent 

K A 
(x 10~ 5 ) 

























B. Further Experiments (Optional) 

2. Measure the pR of a salt such as NH 4 C1 in the concentration range 
0.3-0.01 M and calculate K ion for the weak electrolyte produced in the 
hydrolysis. Compare pH to the calculated pH. 

3. To 100 ml of one of the HAc solutions used above, add a weighed 
amount of NaAc and measure the /?H of the buffer solution. Compare it 
to the calculated value. Account for any difference. 


1. F. R. Richardson, /. Chem. Educ., 33, 517 (1956). 


If one is able to find the molar concentration, X, of fluoride ion in a 
standard CaF 2 solution, it follows that the solubility product constant can 
be determined by 

[Ca+ 2 ][F-] 2 = (X/2)(X) 2 = X 3 /2 

Fluoride may be determined to an accuracy of 0.1 ppm in the range 


in the two flames and to get product colors corresponding to various 
oxidation states of the metal. These colors sometimes are different 
when the bead is hot or cold, and hence a number of ways of differentiation 
of metallic ions is available with one simple method. This finds use in 
the geological field, since many minerals will respond to fusions, whereas 
they are not easily solubilized for wet chemical reactions. 


reducing Jlanoe- 


FIG. 23-1. Making a bead test. Sample and flux are transferred from a watch glass 
via Pt wire to the Bunsen flame. 

Several minerals or salts should be tried as follows: touch a little of the 
material to be tested to the hot, clear bead and heat in the appropriate 
flame until the fusion mixture appears homogeneous. If minerals are 
not available, sulfides like CoS, as prepared in the cation procedures, or 
oxides like Cr 2 O 3 from the side shelf or stockroom may be used with 
good results. See Table 23-3. 

Other compounds may be used instead of borax for fusions, two common 
ones being soda ash, Na 2 CO 3 , and microcosmic salt, NaNH 4 (HPO 4 ). 
Colors developed with several less familiar elements are listed below. A 
typical formula for a compound formed in this method is NaCoPO 4 
(blue). (Color abbreviations refer to A27.) The student may gather data 
to fill in the blanks of Tables 23-3 and 23-4. 

2. Blowpipe Analysis Charcoal Reductions 

(a) Minerals. As usually practiced, one mixes 50 to 100 mg of powdered 
mineral with 100 to 200 mg of Na 2 CO 3 and a drop or two of water. 
This paste is put in a shallow depression dug in a charcoal block. A 
reducing flame directed from the Bunsen flame via a blowpipe fuses the 



Borax Bead Color 

Oxidizing Flame 

Reducing Flame 






Oxide or Metal 




Cr0 3 


Cr a 3 




U0 3 


U 2 3 




Fe 2 3 














Cu 2 O, Cu 




Mn 2 O 3 










Ag, Zn, 
Mg, Si, 
Al, Ca, 



SiO 2 , etc. 



SiO 2 , etc. 


i Microcosmic Salt Bead Color 






















mass. As the mixture tends to melt into the block, more carbonate is 
added and melted as before. This acts as a fluxing material for the 
gangue, leaving the metal oxide to be reduced by the combined action of 
the reducing flame and hot carbon. The result with reducible oxides is 
beads of metal suitable for microscopic or other examination (after 
washing them free of extraneous matter). Powdered minerals or appro- 
priate oxides from the stock shelf may be used. Further tests to identify 
the metals are made after preliminary study with the microscope. Some 
simple cases are listed in Table 23-5. See also Chapter 21, on dry 


Stained ] C Ior Further Test on Free Metal to Identify It 

Ag W Soluble HNO 3 ; precipitates with HC1. (AgCl) 

Au Y Insoluble HNO 3 

Bi Gy Brittle 

Cu R Soluble HNO 3 ; deep blue with NH 4 OH. [Cu(NH 3 )+ 2 ] 

Pb Gy Soft 

Sn W White precipitate with HNO 3 . (H 2 SnO 3 ) 

(b) Sulfur-containing radicals. A specific application and variation of 
the technique is this blowpipe method for sulfur-containing compounds. 
Concentrate a solution suspected of containing S~ 2 , SO 3 2 , S 2 O 3 2 , SCN~, 
or SOj 2 (the last particularly, since good tests for sulfate are less numerous 
than for the others). Heat several drops of this with a little Na 2 CO 3 on a 
charcoal block, directing the reducing flame to the mixture with a blowpipe. 
When well fused to a bead, let it cool, then transfer the bead to a spotplate 
and pulverize it with a stirring rod. If sulfur was present in the unknown, 
sulfide is now present in the powder and any sensitive method for S~ 2 is 
usable. The student should try several of his own design and report the 
results. One commonly used is to put the powder on a shiny, cleaned 
copper or silver coin with a drop of 1 N HC1; a dark spot indicates 
sulfide. The test should be repeated using a precipitate of BaSO 4 in 
place of test solution. In this manner one confirms the barium precipitate 
for sulfate. The reactions are: 

SO^ 2 + 2C = S- 2 + 2C0 2 
S- 2 + 2H 2 = 2OH- + H 2 S 
H 2 S + Cu = CuS B + 2H+ 

3. Examinations With Ultraviolet Light 

(a) Many minerals emit light upon excitation by ultraviolet radiation 
("black light") and are segregated by this means for mining, testing, or 



actual identification. If light is emitted only while ultraviolet radiation 
is falling on the sample, the sample is said to be fluorescent', if emitted 
after the radiation source is removed, the sample is phosphorescent. The 
first phenomena are more generally observed, but the processes are similar 
in mechanism. Other means of excitation are also usable but less common 
in the field. These include heat (thermoluminescence), visible light 
(photoluminescence), X-rays (X-ray fluorescence), sound waves, neutron 
bombardment, etc. Chemicals produced for their fluorescence and 
phosphorescence are used in fluorescent lights, television tubes, advertising 
paints, etc. Some various commercially-coated tubes, minerals, semi- 
precious stones (sometimes identifiable from imitations by this means) 
and chemicals should be studied under ultraviolet illumination and a 
tabulation like Table 23-6 made. Fairly inexpensive ultraviolet lights may 
be purchased, or hand made ones, employing an argon bulb and filter, 
constructed. The former are available in so-called short wave length 
(2537 A) and long wave length (3660 A). The short is used for mineral 
inspection, the long for petroleum investigations. 



Mineral Formula 

Color under 
Natural Light 

Color under 
2537 A 


CaCO 3 




CaWO 4 




SiO 2 -nH 2 O 




A1 2 O 3 


7-0, R 


2ZnO 2 -SiO 2 




CaMg(U0 2 )(C0 3 ) 3 .12H 2 




CaF-Ca 3 Mn(PO 4 ) 3 




CaF 2 








(CaC0 3 .3CaAl 2 Si 2 8 V 




(b) Sometimes fusions are made with minerals to enhance fluorescence 
for the identification of small amounts. The following field test for 
uranium is illustrative of this method. A mixture of 20 parts of NaF, 
400 of Na 2 CO 3 , and 400 of K 2 CO 3 is fused (m.p. 730 C) in a small nickel 
or iron dish to give a glass 1-3 mm deep. (In the field, beer bottle caps 
function as reaction vessels but these are perhaps not as familiar to 
chemists as to geologists.) When cool, one sprinkles a few grains of 
mineral to be tested on the glassy surface, spacing the grains a few milli- 
meters apart. The mass is then reheated until the mineral particles just 
begin to fuse with the flux, then it is allowed to cool. The surface is 
examined under the ultraviolet lamp, using the short wave length filter. 
If the grains contain U, a yellow fluorescence is noted. Some rare earths 
like cerium and a few metals like vanadium if present in large amounts 
will quench the fluorescence, but generally speaking this test gives reliable 
and rapid results. A blank run should also be made, using the reagents 
only. If many samples are to be run the fluxes are prepared in the lab 
and the finished fusions are labeled and brought back from the field for 
closer examination. Alternately, a flux of 10% NaF 90% NaHCO 3 
may be used. 

(c) If a Geiger counter is available its sensitivity should be compared 
with the above test, using for the fusion only an amount of uranium 
mineral which will just give a good fluorescent reaction. It will be found 
that the fusion-ultraviolet test will out perform the usual laboratory 
radioactivity demonstration instrument. 

4. Crystal Types 

The interested student with some background in geology should review 
the section in Chapter 13 on crystals and obtain a standard reference work 
on minerology like Dana's Manual With a hand lens study some well- 
defined chemical compounds and try to identify the crystal symmetry. 
Refer to a chemical handbook for confirmation of decisions. Study some 
minerals in the same way and refer to the handbook on minerology. Note 
in this the variations of the basic types of crystal symmetry and how the 
geologist is able to use them in connection with other simple observations 
like color, hardness, etc. in practical quality analysis. Conventional 
chemical methods do not produce results as rapidly. 


1. J. DeMent, /. Chem. Educ., 23, 213 (1946). 

2. S. M. Edelstein, /. Chem. Educ., 26, 126 (1949). 

3. M. A. Northrup, Anal. Chem., 17, 664 (1945). 

4. E. S. Dana and C. S. Hurlbut, Manual of Mineralogy, 16th ed., John Wiley and 
Sons, New York (1953). 


5. E. S. Dana and C. S. Hurlbut, Minerals and How to Study Them, 3rd ed., John 
Wiley and Sons, New York (1949). 

6. O. C. Smith, Identification and Qualitative Chemical Analysis of Minerals, Van 
Nostrand, New York (1946). 

7. E. E. Bugbee, A Textbook of Fire Assaying, John Wiley and Sons, New York (1940). 

1. Flame Tests 

(a) These tests are used to aid the detection of Cu+ 2 , Sn+ 2 , Ba+ 2 , 
Ca+ 2 , Sr+ 2 , Na+, K+, Li+, and BOj among the ions listed in this text for 
testing. Such tests are usually made by concentrating 1-2 ml of solution 
to about one quarter of its original volume, adding several drops of concen- 
trated HC1, dipping in a clean platinum or nichrome wire loop, and 
holding it in the flame and observing the color imparted to the Bunsen 
flame. (Tests for tin and borate are modifications and are described 
earlier with other examination of these ions.) In order to give a longer 
duration flame one may wind the wire around a little Gooch asbestos 
fiber and moisten with the solution. One should note the color of the 
flame, the time elapsed before the color becomes apparent, and the dura- 
tion of the color. Sodium should be tested last since its flame is jnost 

V Bu 

Gr Y 


R - Color seen 

Y Or 

1 1 

R V 



Gr * Color absort 


4000 4500 

5000 5500 6000 

6500 7000 A 

f~- Ultraviolet 

Infrared * 

FIG. 23-2. The visible spectrum. See also Fig. 3-1. 

Using Pt wire or wire-asbestos, run flame tests on test solutions of the 
ions listed above that have not been done previously. If solutions of 
Rb + and Cs+ are available, add these and any others of interest. Report 
results in the notebook. 

(b) Brilliant flames can be produced by spraying atomized solutions 
(hand bulb or compressed air) into a bunsen flame (HOOD!). 

(c) Potassium in the presence of sodium can be detected upon filtering 
the mixed flame color by viewing it through a sheet or two of cobalt 
glass, the violet-blue glass absorbing the yellow Na+ color and trans- 
mitting the K+ color as a reddish violet. Suggest ways in which Li+ 



could be seen in the presence of Na+ and how a small amount of Na+ 
might be seen in the presence of a large amount of K + or Li+ or Rb+. 
If available, a Di no. 512 filter from Corning Glass Co. is excellent to 
eliminate the Na+ flame so others may be seen. 

2. The Spectroscope 

This instrument is used to make refined observations on flame tests. 
A glass prism separates the light by refraction into its different wavelength 

FIG. 23-3. A spectroscope. The cover has been removed to show the prism. A 

sample in solution is introduced into the flame on a Pt wire loop. The telescope is 

moved to scan the scale upon which are superimposed the spectra. 

components which appear as colored lines called spectral lines. The 
whole visible field may reveal a number of lines for a given heated element, 
compound, or ion and this spectrum is distinctive of the element and 
consequently is a means of qualitative analysis. Each monochromatic 
image of the slit appears to the observer as a narrow, vertical line super- 
imposed on an illuminated scale. The scale may read in wavelengths, 
but more often laboratory demonstration instruments contain an arbitrary 



scale. This is calibrated using known substances in the flame and com- 
paring the arbitrary scale numbers corresponding to the position of their 
strong spectral lines with handbook data that lists actual spectral position 
of these lines. One such instrument containing a scale reading from 
to 180 in divisions of one unit gave the typical data in Table 23-7, 
using a Bunsen burner and the simple method described in part 1. 


Chloride of 

Scale Observation (0-180) 






4.4, 112 



Ca+ 2 



Sr+ 2 

13.5, 26 


Ba+ 2 


Cu+ 2 


The wavelengths corresponding to some of these readings are listed in 
Table 23-8. 

The use of the spectroscope described above was to study emission 
spectra. Another use is the analysis of absorption spectra, which are also 
characteristic of the substances involved. If a beam of white light is 
the source of illumination before the slit in Fig. 23-3, the observer sees a 
continuous ("rainbow") spectrum. If, interposed between the light and 
the slit, is a thick gas layer, a crystal, or a column of solution, the observer 
now sees a discontinuous spectrum, certain black lines appearing since 
energy of those wavelengths, typical of the interposed substance, is being 
absorbed. In general, these dark lines appear where bright, colored lines 
would show if the interposed material had been burned to produce its 
emission spectra. (Using a powerful telescope and refined spectroscope 
in this manner, spectra of the sun have been observed and the composition 
of the absorbing gas envelope around it deduced.) 




Prominent Lines, A 



3274, 3248 

Color due to compound used, not ion itself 


3262, 3175 

Color due to compound used, not ion itself 


5536, 4554 

Color due to oxide, not ion itself 


4227, 3968, 3933 

Color due to oxide, not ion itself 


4607, 4078 

Color due to oxide, not ion itself 


5896, 5890 

Color due to ion itself 


4047, 4044, 7665 

Color due to ion itself 


6707, 6103, 3233 

Color due to ion itself 


7800, 4202 

Color due to ion itself 


4593, 4555 

Color due to ion itself 

Interpose solutions of some colored salts like CuSO 4 , K 2 CrO 4 , CrCl 3 , 
etc. between a white light source and the slit. Record the observations. 
Could unknown samples be identified in this manner if the behaviors of 
known materials were previously catalogued? 

Look up descriptions of the spectrograph, flame photometer, and spectro- 
meter and give schematic sketches of them with brief descriptions. Show 
their similarity and difference with the simple spectroscope. 


1. R. B. Hahn, J. Chem. Educ., 27, 597 (1950). (Na filter) 

2. J. A. Brown, /. Chem. Educ., 30, 363 (1953). (Color filters) 

3. D. M. Smith, Visual Lines for Spectroscopic Analysis, Hilger and Watts, London 

4. L. H. Ahrens, Spectrochemical Analysis, Addison- Wesley, Cambridge (1950). 

5. W. R. Brode, Chemical Spectroscopy, John Wiley and Sons, New York (1943). 


A commercial laboratory advertising itself as performing analyses 
and/or duplicating formulations receives a variety of samples for both 


qualitative and quantitative examination. While such jobs are generally 
not as lucrative as the long-term contracts for routine analyses, the little 
diversified investigations are compensated for by the interest they create 
and the * 'know-how" background the chemist derives from them. 
Examples of these are the type wherein a customer wants to test his 
swimming pool water for free chlorine, another wants to check if a certain 
filling station is actually selling him gasoline containing tetraethyl lead, 
another person has read that coffee contains potassium and is curious 
enough to find out if this is true, a woman wants to verify an advertise- 
ment that a certain beer contains no sugars, another woman wants to 
find out if the dress she bought is wool as purported, or a cotton imitation, 
still another desires to know if an antiperspirant contains metals like zinc 
or aluminum toward which she is allergic, etc. The following six applica- 
tions of qualitative analysis treat the problems just mentioned. 

1. Free Chlorine in Water 

This test is used to test for "free" or "residual" chlorine as found in 
municipally treated drinking water, swimming pools, bottle rinsing solu- 
tions, etc. The usual C1 2 source is sodium hypochlorite or elemental 
chlorine. Other oxidizing agents like Fe^ 3 , Mn lv , and NO 2 interfere 
with the test by oxidizing the reagent and changing its color (see example 
12, Chapter 12, p. 196), but they are not likely to be present in significant 
quantity in ordinary water samples. 

Dissolve 14 mg of o-tolidine, or an equivalent quantity of its dihydro- 
chloride, in 5 ml of H 2 O and add this, with stirring, to a mixture of 3.5 ml 
of H 2 O and 1.5 ml of concentrated HC1. 

To 5 ml of water to be tested and contained in a 4-inch test tube, add 
5 drops of reagent, mix well and allow to stand in the dark for 2 minutes. 
Examine it at that time for a yellow to red color. When running samples, 
one should always compare them with a blank made with distilled water, 
and viewing should be vertically through the longest column of solution. 

The method can be made quantitative by comparison to color standards 
made from Cr vr solutions. See reference 7, p. 436. 

2. Tetraethyl Lead in Gasoline 

(C 2 H 5 ) 4 Pb has been used in gasoline for a long time to moderate the 
combustion of hydrocarbons, thus decreasing motor knock due to too 
rapidly exploding fuel. 

Extinguish all flames in the vicinity. One milliliter of gasoline is 
shaken with 100 mg of decolorizing carbon, centrifuged, and retreated 
if not colorless. A few drops of colorless centrate are put on a spot plate 
or filter paper and allowed to evaporate in the sun. 


While this is taking place, prepare a solution of dithizone (see example 2, 
Chapter 12) by dissolving one crystal in a few milliliters of cp CC1 4 
(redistilled over NaOH preferably) to yield a moderately dark green 

Spot the residue with a drop or two of dithizone. The appearance of a 
red color indicates lead in concentrations as low as 0.02 ppm. The color 
intensity gives an estimate of concentration. (Zinc gives the same 

Lead may also be determined quantitatively in gasoline by precipitating 
it as PbCrO 4 , dissolving in H 2 SO 4 , reacting the Cr 2 O;j- 2 formed with excess 
KI, and titrating the Ij liberated with standard thiosulfate. What are 
the reactions? 

3. Potassium in Coffee 

Potassium is contained in coffee in easily measurable amounts. In 
the human system potassium is needed to activate certain heart muscles. 

Put a gram of ground or powdered coffee in 20-25 ml of H 2 O and 
boil it well. Divide the solution between two tubes, centrifuge, recombine 
centrates, and boil with 200-400 mg of decolorizing carbon. Stir to 
prevent superheating and bumping. Centrifuge and retreat the centrate 
with carbon if it has not been rendered colorless or light yellow. Boil the 
solution down to 2-3 ml and centrifuge again if any sediment remains 
suspended. To 1 ml of cooled, concentrated centrate, add 2 drops of 
1 N HNO 3 and 10 drops of Na 3 [Co(NO 2 ) 6 ]. Mix and note the slow 
formation of yellow dipotassium sodium hexanitrocobaltate, indicative of 
a sample containing K+. The method can be made quantitative by 
centrifuging and reacting the residue with K 2 Cr 2 O 7 + H 2 SO 4 . The com- 
pound reduces dichromate to Cr+ 3 and the green color of the latter is 
matched with a series of standards. 

4. Reducing Sugars in Beer 

In the production of beer, starch (A) in the fermenting malt (wort) is 
hydrolyzed by water and the enzyme diastase to the disaccharide maltose 
(#), which in turn is hydrolyzed in water by the catalytic action of another 
enzyme, maltase, to yield the simple sugar dextrose, also called glucose (C) : 

22 O u - *C 6 H 12 O 6 

(A) (X) (C) 

Beers contain 0.9-2.6 g of sugar, expressed as maltose, per 100 ml. Both 
these sugars are classed as reducing sugars; that is, they are capable of 
working reductions (not on the waistline, however) and become oxidized in 
the process. Both contain aldehyde groups which readily oxidize to acids. 


With Fehling's solution, a basic copper tartrate complex, Cu 11 , is 
reduced to Cu 1 , which is not complexed and precipitates in the hot basic 
solution as red Cu 2 O. The appearance of this precipitate is indicative of a 
reducing sugar in this case, and the method is quantitative if one weighs 
the sample, uses excess reagent, and weighs the cuprous oxide. The 
generalized reaction is 

,0 O 

50H- + 2Cu 1] + R C<f = Cu 2 + R C + 3H 2 O 

Mix 2ml each of Fehling's solutions A ana B (see reagents, A 13) 
and stir to dissolve any Cu(OH) 2 temporarily formed. Now add 2 ml 
of beer and let the tube stand in a hot-water bath for 5-20 minutes. At 
the same time run a known, using a dilute glucose (but not sucrose) 
solution and a blank, using distilled water. The tubes (labeled) are best 
observed while being simultaneously heated in the same bath. 

5. Vegetable and Animal Fibers Compared 

Vegetable fibers like cotton are composed of only C, H, and O, whereas 
animal fibers !k<. wool contain S and N as well. A sensitive test for 
SH groups .a animal protein (as well as sulfur in the form of S~ 2 , 
S 2 O 3 2 , and SCN~) is a sulfur-catalyzed reaction between iodine and 
azide ion A positive reaction gives nitrogen gas bubbles which are 
observable in small concentrations: 

2N 3 - + I 2 = 3N 2 + 21- 

In one spot plate depression put a clean wool thread, in another a 
clean cotton thread. Add to each, 2 drops of H 2 O and 3 drops of 
NaN 3 1 2 reagent (reagent list, appendix), and after a minute or two 
observe which fiber has a cluster of bubbles around it. (The cotton may 
become dye i by the I 2 .) 

6. Aluminium and Zinc in Antiperspirants 

These metallic ions are used in antiperspirant cosmetics because of 
their ability to close skin pores and prevent excretion. 

Heat to boiling about 200 mg of sample with 6 drops of 6 M HC1 and 
3-4 ml oi H 2 O. Stir well, squashing out any greasy lumps to free soluble 
salts. Cool and centrifuge or decant. Divide the clear solution between 
two tubt^ s 

Tube 1 Add 1 drop of phenolphthalein, then 3 M NH 4 OH until barely 
pink, \tiii 1 drop of glacial HAc and rinse the solution into a small 
beaker wuh 3-4 ml of H 2 O, and heat in a water bath to 80 C. Add 1 ml 


of 8-hydroxyquinoline solution (see example 4, Chapter 12 and A 15). 
Then add 2 ml of 2 M NH 4 Ac and keep the mixture hot for 2 minutes. 
A white flocculent precipitate indicates A1+ 3 , Zn+ 2 , and/or Mg+ 2 among 
the common metals precipitating under these conditions. Further testing 
is necessary to show which is present, but a positive test at least indicates 
that further work is needed and also illustrates the use of this important 
organic reagent. 

Tube 2: The student is invited to devise his own procedure for the 
detection of A1+ 3 and/or Zn +2 . Such might be based upon the fact that 
one gives a precipitate with excess NH 4 OH, whereas the other forms a 
soluble ammine complex, etc. Report the results. 


1. S. Weiner, J. Chem. Educ., 26, 318 (1949). (Industrial analysis) 

2. S. P. Nickerson, /. Chem. Educ., 31, 560 (1954). (Pb(C 2 H 5 ) 4 ) 

3. F. D. Snell and F. M. Biffin, Commercial Methods of Analysis. McGraw-Hill, 
New York (1944). 

4. R. S. Young, Industrial Inorganic Analysis, John Wiley and Sons, New York (1953). 

5. Official and Tentative Methods of Analysis of the Association of Official Agricultural 
Chemists, 7th ed., Association of Official Agricultural Chemists, Washington, D. C. 

6. W. W. Scott and N. H. Furman, Standard Methods of Chemical Analysis, Van 
Nostrand, New York (1939). 

7. Standard Methods for the Examination of Water and Sewage, I Oth ed., American 
Public Health Association, New York (1955). 


The scientific criminologist must have a broad background and working 
knowledge of many subjects, and because of the importance and conse- 
quence of his court testimony, he must be an exceedingly cautious and 
painstaking worker. He is frequently called upon to identify and test 
synthetic and inorganic substances as well as materials of plant and animal 
origin and must be versed not only in chemical analysis, but also in 
physics, biology, botany, pharmacology, toxicology, geology, microscopy, 
etc. It is a fascinating application of qualitative testing. The following 
tests are illustrative. 

1. Testing Inks 

Ink types, similarities, differences, composition, ages, etc. may some- 
times be determined by a combination of chemical analysis for the elements 
present and a microscopical examination of the surface appearance of the 


ink and paper. Such work would be likely on questioned documents of 
the type where erasures, fill-ins, additions, counterfeiting, etc. are suspected. 
Thus although two samples of writing may appear alike, one may fluoresce 
under ultraviolet light because of the dye used, and the other may not. 
Or in another case, one ink may contain an iron pigment, and the second 
specimen may contain a copper pigment. Since earliest times inks have 
been compounded and used and thousands of mixtures made. It is 
quickest to demonstrate some differences in two samples as a means of 
telling them apart and qualitative testing is the tool for doing this. 

(a) Testing for iron and the age of inks. Most older inks were made 
from mixtures of ferrous sulfate and two organic weak acids, tannic and 
gallic, with which iron gives a color-fast blue complex. Most newer inks 
are made from organic dyes, glycerine, gums, and water. The older the 
writing with iron inks, the more difficult it is to bleach, thus analyses for 
iron and for bleaching time will aid in characterizing a specimen of 

Put a small piece of paper containing some writing to be examined on a 
watch glass and submerse it in 1 % NaOCl. Note the length of time 
needed to bleach the writing. If writing samples of several ages are 
available (as one might collect from old letters and envelopes) the bleaching 
times should be measured and compared. Experimental procedures, 
including temperature, should be identical. 

The hypochloritc treatment (basic) will cause Fe 43 to be precipitated 
as Fe(OH) 3 where the writing was. If the paper is then washed with a 5 % 
Na 2 S 2 O 3 solution, and next with H 2 O, one may test for the presence of 
iron. The faded writing is spotted with a solution made by dissolving 
0.5 g of NH 4 SCN in 5 ml of 2 M HC1. Red Fe(SCN)+ 2 is the evidence 
one looks for. Blank paper should be run in the same way. Why? 

(b) Testing for chloride. Prepare some AgNO 2 by adding 8 ml of Ag f 
test solution to 3 ml of NO 2 test solution. Decant and discard the 
supernatant liquid, wash the residue with a little H 2 O, and discard the 
liquid again. Dissolve the light yellow residue of silver nitrite in 10 ml 
of3MHN0 3 . 

Cut out a small sample of writing, place it on a watch glass and sub- 
merse it in the silver reagent just prepared. Any chloride in the ink will be 
converted to AgCl and fixed on the paper where the writing was. (Some 
cheaper papers may themselves contain chloride and may mask the next 
reaction.) When the writing has faded, wash the paper with H 2 O, let it 
soak 5 minutes in 0. 1 M HNO 3 , then rinse it with H 2 O again. Now reduce 
the AgCl to metallic silver by spotting the writing (or immersing the 
paper) with a solution made by mixing 5 drops of 35% formaldehyde 
(formalin) in 2.5 ml of 0.5 M NaOH. Again wash the paper in H 2 O 


and if Cl~ was present in the ink, the writing is visible as a tracing of dark 

xO ,0 

H CC + 2O H- + AgCl = H Of + H 2 O + Cl~ + Ag 
\H \0- 

Blank paper is tested in the same way. 

2. Testing Marihuana 

A cooperative police department may furnish a few grams of this common 
narcotic or at least enough for one or two tests to be viewed by larger 
groups. Pictures of the growing plants may also be available. The inter- 
national problem of stopping the illicit narcotics trade has been attacked 
in various ways, not the least important of which is the work of analytical 
chemists whose methods enable law enforcement agencies to test and 
identify microscopic amounts of such materials. While marihuana 
(Indian hemp, cannibus saliva) docs not give the addict the violent with- 
drawal symptoms that opium derivatives do, it is a frequent precursor to 
the use of more powerful narcotics and its insidious character in this 
respect is not to be underestimated. 

(a) Microscopical examination. Under low magnification one looks for 
oval-shaped seeds of light-brown color with tracings in darker brown 
over the surface. Leaf particles will show a surface covered with curved 
pubescent hairs and a few patches of dark resin and pimple-like excre- 
scences called globule protuberances. These features are highly charac- 
teristic of the plant. 

(b) The hydrochloric acid test. Under the microscope, find a globule on 
a leaf surface and pick it apart with a needle. In many of these are found 
crystals of CaCO 3 , hence the name "stone cells" is given to them. Add 
from a capillary glass pipet, made by pulling out a piece of tubing, a 
micro drop of dilute HC1 and note the fizzing as CO 2 is released. 

(c) The Ducanois (Megm) test. Make a solution of 0.2 g of vanillin, 
3 drops of acetaldehyde, and 10 ml of 95 methyl alcohol. Grind a bit 
of marihuana on a spot plate with a blunt stirring rod. Add a few drops 
of petroleum ether and grind again; then let the solvent evaporate. 
Scrape out the larger particles and discard, leaving the extraction stain 
intact. Add 2 drops of 6 M HC1 to this residue, then 4 drops of reagent. 
A positive reaction, which apparently is specific for marihuana, is the 
development over the next 5-20 minutes of a series of colors predominantly 
blue and violet. 

If the three above tests were all positive, the material examined was 
marihuana, and evidence based on similar procedures is being presented 
daily in courts. If other dried plants are available, they % should be 


studied as outlined and differences noted. Among those which show 
some microscopical similarities is catnip. 

3. Testing Blood 

In testing spots, stains, solutions, etc. for blood, one first covers all the 
suspected areas with rapid preliminary tests, then restudies those pieces of 
evidence that gave a positive test with methods that are specific, but more 
time consuming. 

(a) The benzidine test a preliminary. Mix about 0.1 g of benzidine 
base (the free diamine) with 2 ml of glacial HAc and stir well. Add 
4 drops of 3 % H 2 O 2 and stir again. To a speck of material suspected of 
being or containing blood and placed on a spot plate, add a few drops of 
reagent. The appearance of a blue-green color is a positive reaction and 
will detect as little blood as 1-2 ppm. The reaction will even work on 
dried blood samples several years old. Try the test on various samples. 
Some suggested ones are: dried nonhuman blood, rust, MnO 2 , PbO 2 , 
NaOCl, NaCl, NaNO 3 , cloth, perspiration, and hair. Report the results. 
Refer to example 8, Chapter 12. 

(b) The Teichmann test a confirmation. A reagent is made from 0. 1 g 
each of KI, KBr, and KC1 in 10 ml of glacial HAc. A smear of water- 
loosened material (as a stain soaked out of a cloth) is put on a glass slide 
and dried below 100 C. Two drops of reagent are added to the smear, a 
cover glass put on, and the preparation warmed over a low flame until the 
liquid begins to bubble. Then the flame is turned off. Another drop of 
reagent is put alongside the cover slip and allowed to seep under. After 
allowing the slide to cool, examination under about 100 x magnification 
will reveal yellow to brown rhomohedra of hemin chloride. If both tests 
(preliminary and confirmatory) were positive, the sample is blood. 

Further testing is possible, the biological precipitin test verifying the 
sample to be human blood and giving its type: A, B, AB, or O. Experts 
in the field are able to group samples even more specifically and some 
feel that in time enough will be known about blood and its analysis so 
that blood samples will become as distinctive as fingerprints in human 


1. C A. Schmitt, J. Chem. Educ., 21, 413 (1944). 

2. S. H. Roberts, J. Chem. Educ., 22, 245 (1945). 

3. L. Jones, R. Pinker, and W. Penprase, Los Angeles Police Dept. Crime Laboratory, 
private communications. 

4. P. L. Kirk, Crime Investigation, Interscience, New York (1953). 

5. S. Smith and W. G. H. Cook, Taylor's Principles and Practice of Medical Juris- 
prudence, 10th ed., Churchill, London (1948). 


6. R. B. H. Gradwoh), Clinical Laboratory Methods and Diagnosis, 3rd ed., Mosby, 
St. Louis (1943). 

7. H. T. F. Rhodes, Forensic Chemistry, 2nd ed., revised, Chapman and Hall, 
London (1946). 


1. Boron 

A test for boron which can detect that element at low concentrations is 
reported by White. The method involves making a derivative of boron 
(as borate) with benzoin and checking the fluorescence under the ultra- 
violet light. 

Prepare 1 ml samples, using borax as a source of boron, to contain 
0.5, 5.0, 10, 50, and 100 ppm B. To each add 1 drop of 0.25 M NaOH, 
then 8 ml of 95% ethyl alcohol, and 1 ml of 0.4% benzoin in alcohol. 
Shake each and compare the strengths of fluorescence in a dark room with 
ultraviolet radiation. A green-white color is specific for boron. Does it 
appear from these samples that one could make the method semi- 

2. Aluminium 

A similar method is usable for metallic ions that yield 8-hydroxyquino- 
line (oxine) derivatives. (See example 4, Chapter 12, and special experi- 
ment 6, part 6.) 

Put 2 drops of Al 43 test solution on a piece of filter paper, add 2 drops 
of an alcoholic 8-hydroxyquinoline solution, and show that the spot 
fluoresces only weakly. Now hold the paper over an open NH 4 OH 
bottle. Reaction takes place between the oxine and metallic ion and the 
spot becomes strongly fluorescent. Wash the spot with a little H 2 O, 
then hold over an open bottle of glacial HAc. The spot is still fluorescent, 
showing the stability of the Al-oxine chelate. 

Repeat the experiment with various quantities of the metal and note the 
effect on the fluorescence. 

Repeat with Mg+ 2 instead of Al 13 . What difference is noted? Does 
this give one a method for determining which chelate is more stable? 
Postulate a reason for the stability differences. 


1. A. L. Powell, J. Chem. Educ. t 24, 423 (1947). 

2. C. E. White, J. Chem. Educ., 28, 369 (1951). 

3. F. Feigl and G. B. Heisig, /. Chem. Educ., 29, 192 (1952). 


4. H. C. Dake and J. DeMent, Ultraviolet Light and Its Applications, Chemical 
Publishing, New York (1942). 

5. J. DeMent, Fluorescent Chemicals and Their Applications, Chemical Publishing, 
New York (1942). 


Chromatography is a method of separating one or several dissolved 
components of a mixture by distributing them between two immiscible 
phases, one of which is moving past the other. Separations can be made 
in columns (Twsett, 1906) packed with insoluble materials like alumina, 
silica, starch, carbon, etc. over which the solution is poured, or made on 
paper strips (Goppelsroeder, 1906) over which the solution moves capillarity. 
In either case, under proper conditions the separable components will 
concentrate in different areas and will appear as variously colored zones 
or bands, either by themselves or by color-developing reagents or radia- 
tions. The components may be bound chemically or physically to the 
adsorbing surface and since the various components have different abilities 
to do this, separations are possible which might be difficult by other 
means. Considerable opportunity for original work in this technique 

1. Separation of Two Dyes a Quick Preliminary Illustration 

Dissolve a few crystals of salts of methyl orange and malachite green in 
several milliliters of H 2 O contained in a 7- or 8-inch test tube. Cut a 
strip of filter paper \ in. wide and long enough to reach to the liquid 
from the tube's mouth. Fix the strip at the top by some method, such as 
a staple and cork, so that it hangs free in the center of the tube with its 
bottom about 1 cm below the level of the dye mixture. Observe over 
the next few hours that the methyl orange rises much more readily than 
the green dye and that a separation is effected. In practice one would 
cut the strip in two at the color boundary and elute each component from 
the paper with proper solvents. When the solvent boundary has almost 
reached the top, calculate the R f value (ratio of fronts) for each dye: 
Rj = distance the component traveled on the strip divided by the distance 
the solvent traveled. 

2. Separation by Partition on Paper of Group 1 Cations 

Cut a strip of filter paper about 30cm long and 1.5cm wide. Put 
3 drops of each group 1 test solution on one spot about 5 cm from the 
bottom of the strip and dry it over a low flame. Suspend the strip from 
a cork in a cylinder which contains at the bottom, in contact with the 



strip, a few milliliters of a mixture containing 15% H 2 O, 10% ethylaceto- 
acetate, 75 % -butanol, and enough glacial HAc to make the pH 3.5-4. 
After the liquid boundary has reached the top, remove the strip and let 
it dry. Spray the strip with a dilute K 2 CrO 4 solution and from the colors 
of the group 1 chromates discovered in the preliminary experiments in 
Chapter 15, identify each zone. Calculate R f for each ion. Wash the 
strip with H 2 O then dry it and label its areas. See reference 1, p. 443. 

Many references may be found giving similar methods for the other 
anion and cation groups. The separations may be made quantitative by 
dissolving the adsorbed substance from the paper and then titrating, 
electroplating and weighing, ashing and weighing, etc. 

3. Separation in a Column of Group 1 Cations 

Prepare a dry column as illustrated in Fig. 23-4. Start moderate 
suction with the aspirator, add 2 ml of H 2 O, and press the cotton wad 

I.D. tube 




AI 2 5 , 3Z5 mesh , 
Section ~20 cm long 

glass wool 

4-mrw I.]), tube 

to trap and aspirator 

FIG. 23-4. A chromatographic column. 

down well to keep air moving ahead of the water, excluding entrained 
bubbles from the column. Adjust the aspirator so that the water boundary 
falls quite gradually. Add in order, 1 ml of glacial HAc, 1.5 ml of 5% 
K 2 CrO 4 , 2 ml of H 2 O, 3 ml of solution containing 1 ml of each group 1 
test solution, Ag+, Pb^ 2 , Hg+ 2 , and then 3 ml of H 2 O. 

Bands of colored chromates will form in the column at different distances 
from the top. Note the order and compare it to the paper method. 
The column may be pushed out with a plunger, the colored segments cut 
out with a spatula, the metallic chromates dissolved in HNO 3 , and those 
solutions further tested. Outline a method for testing to verify the 
presence of each metallic ion. See reference 6, p. 443. 


4. Group 2 by Paper Chromatography 

Cut a strip of Whatman no. 1 filter paper about a foot long and i in. 
wide. Put a J in. streak of solution containing several group 2 ions across 
the width of strip and about 1 in. from the bottom. Dry it by waving 
over the Bunsen flame. Hang the strip from a cork in a 12-in. tube which 
contains enough developing solvent (95 volumes of tert-butanol and 
5 volumes of glacial HAc) to submerge about | in. of the strip below the 
streak. Do not get solvent above the streak. Stand the tube aside for 
6-12 hours. Remove the paper when the solvent boundary reaches the 
top, and allow the strip to air or oven dry. Moisten the strip with H 2 O, 
using a fine atomizer or steam bath, and then expose it to H 2 S gas until 
all the colored sulfide zones have appeared. One may cut the individual 
areas out, dissolve them in HC1 or HNO 3 and run spot tests to determine 
their identity, or one may identify them simply by reference reading, then 
dry the strip, label the zones, and preserve the preparation with clear 


1. J, G. Surak and D, P. Schlueter, /. Chem. Edttc.. 29, 144 ( 1952), and 30, 457 (1953). 

2. J, G. Surak, N. Leffler and R. Martmovich, /. Chem. Educ., 30, 20 (1953). 

3. A. Klmkenberg, /. Chem. Educ., 31, 423 (1954). 

4. H. G. Cassidy, Anal. Chem., 24, 1415 (1952). 

5. H. H. Strain and G. W. Murphy, Anal. Chem., 24, 50 (1952). 

6. O. C. Smith, Inorganic Chromatography, Van Nostrand, New York (1953). 

7. R. J. Block, R, LeStrange, and G. Zweig, Paper Chromatography, Academic Press, 
New York (1952). 

8. T. Moran, California Polytechnic at San Dimas, personal conversations. 

9. P. Dickerson, Columbia Records, personal conversations. 


Dithizone (diphenylthiocarbazone) was prepared by the great organic 
chemist Emil Fischer in 1878, but its use as a color reagent for small 
amounts of metallic ions did not become generally known until almost 
50 years later. It is soluble in several common organic solvents and with 
many metal ions forms inner complexes which are also soluble in these 
solvents. If the organic solvent is immiscible with H 2 O, it affords a 
method of extraction, and, since the complexes of the metals form at 
various /?H's, and the colored dithizonates lend themselves to quantitative 
colorimetry, considerable use of the reagent has been evidenced during 
recent years, Dithizone is a sensitive reagent for Mn, Fe, Cu, Ni, Co, 
Zn, Pd, Ag, Cd, In, Sn, Pt, Au, Hg, Tl, Pb, and Bi. Good technique is 
needed, of course, since its wide reactivity makes any specific reaction 
subject to question. See example 2, Chapter 12. 


1. As an Indicator Paper 

Make about 10 ml of a saturated solution of dithizone in acetone. 
Soak a circle of filter paper with the solution and when the solvent 
evaporates, cut the paper into small squares, each suitable for a single 
test. To succeeding squares, add a drop of the following, and record the 
colors produced initially and when dry: 1 M HC1, 1 M NaOH, 1 M 
NH 4 OH, and ion test solutions of Zn, Pb, Sn, Hg 2 , Hg, Cu, Ag, and Bi. 
See reference 2 below. 

2. As a CCI 4 Solution 

The reagent may be prepared as a 0.001 wt/volume per cent solution in 
cp CC1 4 or CHC1 3 . To test an aqueous solution, 2 ml of solution are 
shaken with 10 drops of dithizone reagent and the color in the dithizone 
layer noted. In practice, a buffer is also added to aid specific extractions. 
Working in this way, as little as 0.0016 ppm Zn+ 2 has been reported 
detectable, which is comparable to spectrographic analysis and is one of 
the most sensitive chemical tests known. Water from galvanized pipes 
easily shows this test and distilled water which has been condensed in 
copper tubing gives a test for copper. The reagent is so sensitive in fact 
that, in inexperienced hands, almost any aqueous sample gives a positive 
reaction of some sort. 

Dilute the same test solutions used in paragraph 1 by 10-fold and run 
them as mentioned here, without buffers, and report the colors obtained 
in the CC1 4 layer. 


1. E. B. Sandell, Colorimetric Determination of Traces of Metals, 2nd ed., Inter- 
science, New York (1950). 

2. J. T. Dobbins and J. H. Norman, /. C/iem. Educ., 27, 604 (1950). 


There are many examples of the identification of substances by micro- 
scopical study of the crystal habit of reaction products. This general 
method is used particularly with complex substances like natural products 
which do not yield to other simple tests. One drawback is that a given 
material may, under slightly different conditions of temperature, concen- 
tration, etc., yield differently shaped crystalline products and one may 
misinterpret the evidence, so experience in handling the technique is all 

In group 4, calcium and strontium are often confused because of their 


chemical similarities. The following method is one means of differentia- 
tion and is typical of the use of the microscope in this behalf. See 
reference 3 below. 

1. Clean a glass slide thoroughly with scouring powder, rinse it under 
tap and distilled water, then let it dry. 

2. Near one end of the slide put a drop of Ca+ 2 solution containing 
5-6 mg Ca^ 2 per milliliter and with a clean Pt wire spread the drop a bit. 

3. Near this drop put a drop of 10% HIO 3 and draw the cleaned 
wire from it to the first drop, making a thread of connecting reagent 
between the two. Tilt the slide only slightly to allow the acid to flow. 
This more dense solution will mix on the bottom of the calcium drop and 
with crystallization starting there, crystals will show their well-defined 
features toward the observer. 

4. After the spot has been prepared, it is examined under low power, 
25-1 50 x, and the crystal growth and habit described in the notebook. 

5. Repeat 1-4 using strontium at the same concentration at the other 
end of the slide. Note the differences in the two preparations. Comment 
on this specific method and on the general approach to qualitative analysis. 

6. As an alternate precipitating agent in the same general manner, use 
6 M H 2 SO 4 and warm the slide gently until crystals begin to appear at the 
edges. Note the characteristic bundles of needles exhibited by 
CaSO 4 2H 2 O. 

The use of the microscope illustrated in this experiment is only one of 
the ways in which the analyst utilizes this important instrument. Further 
applications are in finding the index of refraction of small samples (which 
is an identification means) and in observations using polarized light, under 
which crystals show several identifying characteristics. The uses of the 
microscope in metallurgy, criminology, and petrography are invaluable. 


1. W. C. McCrone, Anal Chem., 26, 42 (1954). (Note Bibliography). 

2. L. Levi and C. G. Farmilo, Anal. Chem., 26, 1040 (1954). 

3. E. M. Chamot and C. W. Mason, Handbook of Chemical Microscopy, 2nd ed., 
2 vols., John Wiley and Sons, New York (1938). 

4. H. F. Schaefier, Microscopy for Chemists, Van Nostrand, New York (1953). 


One gram-mole of acetamide (CH 3 CONH 2 , mol wt = 59) and 0.12 g- 
mole of powdered phosphorous pentasulfide (P 4 S 10 , mol. wt = 444) are 
refluxed in a liter flask (steam bath) containing a few boiling chips and 
500 ml of dry benzene, for about a half hour. The reflux column should 
carry a CaCl 2 tube at the top. The solution is decanted from the residue 


into another flask, refluxed with several boiling chips and 5-6 g of 
decolorizing carbon, then filtered hot. The flask is connected to a con* 
denser via a I0-mm I.D. goose neck and excess solvent distilled, using the 
steam bath until crystallization of product appears imminent. The hot 
solution is poured into a beaker and allowed to crystallize (hood) over 
night. The thioacetamide is filtered off on a suction assembly and the 
mother liquor either concentrated to obtain a second, small crystal crop, 
or returned to the original flask with the distilled benzene and residue from 
the first refluxing. This flask is recharged with more solids and solvent 
as needed to give approximately 500 ml, and the process repeated. After 
the final run, the gummy residue in the reaction flask is removed with 
small portions of dilute HNO 3 and later heating. (HOOD! CAUTION!) 
The yield is about 30% of theoretical. The reaction is 

CC + P 4 S 10 = P 4 10 + 10CH 3 -C 

N NH 2 NH 2 

Good ventilation must be provided as moisture will cause some hydrolysis 
liberating hydrogen sulfide. 

P 4 S 10 + 16H 2 O = 10H 2 S + 4H :) PO 4 

The 8 % aqueous student solutions need not be filtered prior to use if a 
residue appears. 


1. H. Leo, Ber. 9 10, 2133 (1877). 

2. A. Bernthscn, Ber. 9 11, 503 (1878). 

3. A. W. Hofmann, Ber., 11, 504 (1878). 

4. A. Hantzch, Ann., 250, 257 (1889), 

Also see references at the end of Chapter 16. 


Another simple and useful technique in qualitative testing is the electro- 
graphic method first described by H. Fritz in 1929. The sample under 
test is anodically eroded onto a piece of filter paper charged with a moist 
inert electrolyte solution like KC1 or K 2 SO 4 . The paper is then spotted 
with a reagent and the sample identified by the color produced. Typical 
reagents are the same ones used in the laboratory to give color reactions 
with the common ions : dimethylglyoxime, silver nitrate, acid-thioaceta- 
mide, ammonium sulfide, potassium ferrocyanide, potassium chromate, 
etc. The apparatus is shown in Fig. 23-5, A desirable addition is a 
means for exerting pressure on the electrode-sample system. 



Any d-c source (from a flashlight battery to a small rectifier) is satisfactory 
and other metals than those indicated may be used for electrodes. The 
anode is a rod, and the cathode is a sheet. The sample is a metal, con- 
ducting mineral, or even powder. The meter is any low current-consuming 


FIG. 23 5, The electrographic setup. Pressure must be put on the anode to obtain 

a good print. Contact is maintained a minute or more For serious work, a regular 

press is used to produce pressures up to IO(K) psi. 

ammeter or voltmeter, but it is not absolutely necessary. A little experi- 
mentation will reveal the conditions of time and solution concentrations 
that give best results with a particular device and sample. As with the 
rest of qualitative testing, the method is open for improvement and 
ingenuity by the person with the scientifically inquisitive mind. 


1. H. Fritz, Z. Anal. Chem., 78, 418 (1929). 

2. R. Wilbur, Pacific Division, Bendix Aviation Corp., personal conversations. 

3. J. J. Lingane, Electroanalytical Chemistry, Interscience, New York (1953). 



1. A. Hybbinette, Anal Chem., 17, 654 (1945). (Minerals by microchemistry) 

2. W. E. McKee and W. F. Hamilton, Anal. Chem., 17, 310 (1945). (HF, HNO 3 in 
pickling baths) 

3. H. Ravner, Anal Chem., 17, 41 (1945). (Cu, Pb, Sn, Mn, Fe, Ni, Al in bronze) 

4. R. T. Phelps, E. A. Gulbransen, and J. W. Hickman, Anal Chem., 18, 391 and 
640 (1946). (Oxide film microscopy) 

5. 0. J. Kelley, B. S. Hunter, and A. J. Sterges, Anal. Chem., 18, 319 (1946). (N, P, 
K, Ca, Mg in plants) 

6. C. E. Danner and J. Goldenson, Anal. Chem., 19, 627 (1947). (Incendiaries and 

7. E. Wichers and others, Anal Chem., 19, 210 (1947). (Impurities in analytical 

8. R. N. Bell, Anal. Chem., 19, 97 (1947). (Phosphoric acids) 

9. J. H. Karchmer and J. W. Dunahoe, Anal. Chem., 20, 915 (1948). (Sulfur com- 
pounds in refinery waste) 

10. J. J. Furey and T. R. Cunningham, Anal. Chem., 20, 563 (1948). (W, Cb, Ta, Ti, 
Co, Ni, Cr, Fe, Si, Mn, C in W alloys) 

11. E. E. Baskerville, Anal. Chem., 21, 1089 (1949). (Cu, Be, Zr, Ni, Co, P, Ag, 
Cr in Cu alloys) 

12. P. J. Elving and P. C. Chao, Anal. Chem., 21, 507 (1949). (Alkali metals in 

13. C. W. Mason, Anal. Chem., 21, 430 (1949). (And four articles following) 

14. S. K. Love, Anal. Chem., 21, 278 (1949). (Water analysis) 

15. B. L. Oesper, Anal. Chem., 21, 216 (1949). (Food analysis) 

16. K. D. Jacob, Anal. Chem., 21, 208 (1949). (Fertilizer analysis) 


Al. Review of Exponential Numbers 

Very large and very small numbers are common in chemistry. They 
are best expressed and mathematically handled by writing them as one 
decimal numbers times 10 to a power (exponent). For transposing 
numbers larger than 10 into this form, one moves the decimal point to the 
left the requisite number of places to still leave one place to the left of the 
point, and the resulting exponent is a positive integer expressing the 
number of places the decimal was moved. Thus 

1 15 = 1.15 x 10 2 (decimal point moved two places) 

6,000,000 = 6.00 x 10 6 (decimal point moved six places) 

3717 = 3.717 x 10 3 (decimal point moved three places) 

One may round off the third figure, as slide rule accuracy is sufficient in 
problem solving in this course. The last number then becomes 3.72 x 10 3 . 
In expressing numbers smaller than 1 in this system, the decimal is 
moved to the right to give a one-decimal number. The exponent again 
expresses the number of places moved but is negative this time: 

0.156 = 1.56 x lO" 1 (decimal point moved one place) 

0.00326 = 3.26 X 10 3 (decimal point moved three places) 

0.000000407 = 4.07 x 10~ 7 (decimal point moved seven places) 

Common mathematical operations are speeded in the following ways. 



(1) Multiplication 

Numbers are written in exponential form as described above. The 
numbers preceding the "times" signs are multiplied on the slide rule (see 
slide rule section) in the normal fashion. The exponents are added 
algebraically. The answer is finally written as a one-decimal number 
times 10 to a power. Thus 

(2 x 10 3 )(4 x 10 4 )(3 x 10 2 ) = 24 x 10 5 = 2.4 x 10 6 (Ans.) 

(2) Division 

The decimal parts are divided on the slide rule in the usual manner. 
The sign of the exponent in the denominator is reversed and the exponents 
in numerator and denominator then added algebraically: 

5.75 X 10- 5 /7.21 x 10 3 = 0.798 X 10~ 8 = 7.98 X 10 9 (Ans.) 

(3) Addition and subtraction 

the numbers are rewritten (it necessary) so that all exponents are alike 
in magnitude and sign. The decimal parts are then added or subtracted 
in normal fashion and the answer has the same exponent on 10 as did all 
the preceding 10's in the problem: 

2.00 x 10 2 + 3.00 x 10 l + 1.00 x 10 3 

= 0.200 x 10 3 + 0.0300 x 10 3 + 1.00 X 10 3 
= 1.23 x 10 3 (Ans.) 

(4) Squaring 

Square the decimal part using the slide rule, then multiply the exponent 
by 2 and rewrite the answer in one decimal exponential form: 

(4000)' (4.00 x JO 3 ) 2 = 16.0 X 10 6 = 1.60 x 10 7 (Ans.) 

(5) Cubing 

Cube the decimal part using the slide rule, then multiply the exponent 
by 3 and tew me the answer in one decimal exponential form: 

(300) :1 = (3.00 X I0*) 3 = 27.0 X I0 6 = 2.70 x 10 7 (Ans.) 

(M Vi/wj/t 1 r*>.;' (the | power) 

Rewrite the numbei so the decimal part is a number between 1 and 100 
and the exponent is divisible by 2. Find the root of the decimal part 
using the slide rule in the usual way, then divide the exponent by 2. The 


answer will always contain a decimal number between 1 and 10 and may be 
left as found : 

^2500 = (2500)* = (25.0 x 10 2 )* = 5.00 X 10 l = 5.00 X 10 (Ans.) 

(7) Cube root (the J power) 

Rewrite the number so that the decimal part is a number between 1 and 
1000 and the exponent is divisible by 3. Find the root of the decimal 
part using the slide rule, then divide the exponent by 3. The answer will 
always contain a decimal number between 1 and 10 (why?) and may be 
left as found : 

^0.000722 = (722 x 10~ 6 )* = 8.98 x 1Q- 2 (Am.) 

A2. Review of the Slide Rule 

The following general rules constitute only the briefest summary of the 
operation of those scales on the slide rule that are necessary for working 
problems encountered in this course. The student is referred to the 
library for books giving more detail and worked examples. -Consult 
these or your instructor regarding questions arising from markings on 
nonstandard slide rules, special scales, etc. It is assumed the student is 
able to read the scales on the rule; i.e., he can locate given numbers and 
can read the answer. It is best if the numbers in the problem be rewritten 
in exponential form (see preceding section), the answer approximated 
mentally (since the slide rule only gives the numbers in the answer and 
does not directly give decimal points), and then the actual answer found 
with the rule. Thus 


= (4.06 x 10 2 )*(2.84 x 10)/(3.16 x 10)(1.50 x 10~ 2 ) 

^ (2 x 10)(3 x 10)/(3 x 10)(1.5 x 10~ 2 ) 

^ 2 X 10/1.5 x 10- 2 ^ 1.3 X 10 3 (Approx. am.) 

The number one gets on the slide rule is 121. The actual answer must 
be 1.21 x 10 3 . The 10-in. slide rule gives 3-place accuracy, and that will 
be sufficient for our purposes. If the student is in doubt as to the inter- 
action of any two scales, he may always check his methods by trying a 
problem (involving those scales) whose answer is known. 


(1) Multiplication 

To multiply x by ?/, put the index of the C scale (the 1's at the ends of the 
scale are the indexes) over x on the D scale. Slide the wire (also called 
the cursor) over ?/ on the C scale, and read the answer under the wire on 
the D scale. In case the answer does not fall on the rule, having used the 
left C index, the right C index is used, thus permitting any number of 
numbers to be multiplied together without running out of room on the 
rule or having to write down any intermediate answers. 

(2) Division 

To divide x by //, (x/y), put wire over x on D scale. Slide y on C scale 
under wire. Read answer under the C index on the D scale. 

(3) Logs ami Antilogs 

(More detail is given in the section following, which is devoted to this 
topic.) Most rules work in this manner: to find the mantissa of the 
logarithm of x, put wire over x on D scale, read the mantissa under wire 
on the L scale. To find the antilog of //, put wire over // on L scale, read 
antilog under wire on D scale. 

(4) Squaring 

To square .r, (.r 2 ), rewrite f x as a one-decimal exponential number, put 
wire over .r on D scale, and read square under wire on A scale; multiply 
the exponent by 2. Note that the A scale is a double one, going from 1 
through 10 to 100. Thus 2 2 is 4, but 6.33 2 is 40.0 and (6.33 x 10 4 ) 2 is 
40.0 x 10or4.00 x 10 9 . 

(5) Cubing 

To cube a:, (.r 3 ), write x as a one-decimal exponential number, put wire 
over x on D scale, and read cube under wire on K scale; multiply exponent 
by 3. Notice that the K runs from 1 through 10 and 100 to 1000 in a 
triple scale. Thus 2 3 is 8, 4.3 1 3 is 80.0, and 9.29 3 is 800. 

(6) Square root 

To take the square root of ;r, (Vx or x), write x as a number between 
1 and 100 times 10 to an even exponent. Put wire over x on A scale, 
read root under wire on D scale; divide exponent by 2. 

V650 = V6.50 x 10 2 = 2.55 x 10; and V6500 = V65.0 x 10 2 
= 8.07 x 10 (Ans.) 

(7) Cube root 

To take the cube root of x, (3/x or x$), write x as a number between 1 



and 1000 times 10 to a power divisible by 3. Put the wire over x on the 
K scale, read root under wire on D scale; divide exponent by 3. 

^5 = 1.71; ^500 = 7.94; 
^50,000,000 = (50.0 x IO 6 )* = 3.69 x IO 2 (Ans.) 

Other scales will find little use in this course. If desired, they can be 
mastered by use of a manual on the slide rule. 

A3. Review of Logarithms 

If one understands the foregoing section on exponential numbers, he 
should have little difficulty with logarithms. In this course, only logs to 
the base 10 will be used. The following development summarizes their 
use. See also A30. 

The logarithm of any number is the power to which 10 must be raised to 
equal that number: 


Number Exponentially 



IO 5 



IO 2 



10 1 





IO- 1 



io- 4 


For numbers like 75, which cannot be expressed as an even power of 
10 and handled mentally, one uses log tables or the L and D scales on the 
slide rule. Obviously the log of 75 must lie between 1 and 2 (IO 1 and IO 2 ) 
and is 1.875. 

Each log is made of two parts: the characteristic, which is to the left 
of the decimal point and establishes the decimal places in the actual 
number the log represents; and the mantissa, which is to the right of the 
point and establishes the integers in the actual number the log represents. 
It is seen above that the characteristic is 1 less than the number of decimal 
places in the actual number. Example: Find the log of 6750. One 
rewrites this as 6.750 x IO 3 . The characteristic is 3 and the log of 6.750 
is found in a table to be 0.829, so the log of 6750 is 3.829. The operation 
is the addition of two logs: log 6.750 + log IO 3 = 0.829 + 3 = 3.829. 
As we added exponents in the previous section on multiplication, so we 
will add logs to carry out the same process. Using most slide rules, one 


vcs the slide wiie over 675 on the D scale and reads 829 under the wne 
on the L scale. (Some models operate differently.) 

The log o( a numbei less than I has a negative value. I ind the log 
ot QOCXP50. I his U lewritten 6,750 x 10 :t = log 6 750 f log 1Q- 3 
~ 0.829 - 3 = - 2 1 71. If one has this log given and is asked to find 
the decimal number it lepresenls, the process is a little more complicated 
Mnce one has to convert 10 - 2 - 171 to the number desired, and log tables 
do not provide for negative mantissae. The mantissa is made positive 
by rewriting as a pioduct of a plus and a minus exponent such that their 
addition gives the negative exponent in question. Thus 10 2 - IT1 
= 10 3 x 10 * 2 ''. This becomes 10 a x antilog 0.829 = 6.750 x 10~ 3 . 

Examples: (a) Using logs find the product (0 021 5)(6 31 x 10 5 )(227). 
= log (2.15 x 10 2 ) -f log (6.31 x I0 5 ) + log (2.27 x 10 2 ) 
= (0.332 - 2) 4- (0.800 -f 5) 4- (0.356 4- 2) 
= - 1.668 + 5.800 + 2.356 = 8.156 - 1.668 = 6.488 
, b.48 ^ | Q 6 x , o.4HH = , 6 x antl | og 0.488 = 3.08 x 10 fi (,4m.) 

(/>) Find JT- r = log (1/2.56 x 10 7 ) 

The problem is simplified slightly by making part of the indicated division 
first, then subtracting the two logs to carry out the next division: 

= log ( IO ? /2.56) = log 10 7 - log 2.56 = 7 - 0.408 = 6.592 ( 4//v.) 

(This is typical of />H problems, Chapter 6.) 
(f) Find r using logs: r= \ 3 001 75 

= 1/3 log (1.75 x 10 *) = 1/3 (0242 ~ 2) 
= 1 758/3 = - 586 
I0-o.ri86 = 10- 1 x IO"- 414 = 10 - l x antilog 0.414 = 2.60 x 10 ( 4m ) 

Checking on the slide rule: does(2.60 x 10 ] ) 3 = 1.75 x 10 2 (D and K scales.) 
Note again the relationship between exponential handling and working with logs. 

Students having slide rules with LL scales may use those to take numbers 
to powers and use logs to check, or vice versa. For instance, 3.80 2 - 7 "': 
set left index of C scale over 3.80 on LL3 scale. Put wire on 275 on C 
scale, and read 39.3 under wire on LL3 scale. How are logs used to 
check this? 

It is handy to learn a few logs of numbers to aid in estimating answers. 
These are easy to remember: log 1 = 0, log 10 = 1, log 2 = 0.3, 
log 5 = 0.7. Other identities follow; as, 2 x 2 = 4, meaning that log 2 
+ log 2 = log 4, and since log 2 = 0.3, then 0.3 + 0.3 = 0.6 =T log 4; etc. 

For practice in using logs and the slide rule, the student should review 
problems in his lecture notes and those worked in the text so he has 
answers to check his computations. 

What was stated previously regarding the handling of exponential 



numbers in multiplication, division, extracting roots, squaring and cubing 
may now be applied to logs. Thus: 

log (A-B) = log A + log 
log (A/B) = log A - log B 

log A x = x-logA 
log \/A = (I/*) log A 

To find the actual number in question after the above operations are 
done, one refers to a table of antilogs or uses his slide rule in reverse 
order to that used before: the mantissa is located under the wire on the 
L scale and the corresponding number (antilog) is read under the wire 
on the D scale. One then places the decimal point as dictated by the 

See appendix A28 for a log table. 











Stirring rod, glass 





Stoppers, assorted 





Test tube